tcu11 11 01 m

Other infinite series do not have a finite sum, as with

The sum of the first few terms gets larger and larger as we add more and more terms. Tak-ing enough terms makes these sums larger than any prechosen constant.

With some infinite series, such as the harmonic series

it is not obvious whether a finite sum exists. It is unclear whether adding more and moreterms gets us closer to some sum, or gives sums that grow without bound.

As we develop the theory of infinite sequences and series, an important applicationgives a method of representing a differentiable function ƒ(x) as an infinite sum of powersof x. With this method we can extend our knowledge of how to evaluate, differentiate, andintegrate polynomials to a class of functions much more general than polynomials. Wealso investigate a method of representing a function as an infinite sum of sine and cosinefunctions. This method will yield a powerful tool to study functions.

1 +12

+13

+14

+15 +

16

+Á

1 + 2 + 3 + 4 + 5 +Á .

INFINITE SEQUENCES

AND SERIES

OVERVIEW While everyone knows how to add together two numbers, or even several,how to add together infinitely many numbers is not so clear. In this chapter we study suchquestions, the subject of the theory of infinite series. Infinite series sometimes have a finitesum, as in

This sum is represented geometrically by the areas of the repeatedly halved unit squareshown here. The areas of the small rectangles add together to give the area of the unit square,which they fill. Adding together more and more terms gets us closer and closer to the total.

12

+14

+18

+116

+Á

= 1.

746

C h a p t e r

11

1/2

1/4

1/8

1/16

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DEFINITION Infinite SequenceAn infinite sequence of numbers is a function whose domain is the set of positiveintegers.

The function associated to the sequence

sends 1 to 2 to and so on. The general behavior of this sequence is de-scribed by the formula

We can equally well make the domain the integers larger than a given number andwe allow sequences of this type also.

The sequence

is described by the formula It can also be described by the simpler formulawhere the index n starts at 6 and increases. To allow such simpler formulas, we

let the first index of the sequence be any integer. In the sequence above, starts with while starts with Order is important. The sequence is not the same asthe sequence

Sequences can be described by writing rules that specify their terms, such as

dn = s -1dn + 1

cn =

n - 1n ,

bn = s -1dn + 1 1n ,

an = 2n ,

2, 1, 3, 4 Á .1, 2, 3, 4 Áb6 .5bn6

a15an6bn = 2n ,

an = 10 + 2n .

12, 14, 16, 18, 20, 22 Á

n0 ,

an = 2n .

a2 = 4,a1 = 2,

2, 4, 6, 8, 10, 12, Á , 2n, Á

Sequences

A sequence is a list of numbers

in a given order. Each of and so on represents a number. These are the terms ofthe sequence. For example the sequence

has first term second term and nth term The integer n is calledthe index of and indicates where occurs in the list. We can think of the sequence

as a function that sends 1 to 2 to 3 to and in general sends the positive integer nto the nth term This leads to the formal definition of a sequence.an .

a3 ,a2 ,a1 ,

a1, a2, a3, Á , an, Á

anan ,an = 2n .a2 = 4a1 = 2,

2, 4, 6, 8, 10, 12, Á , 2n, Á

a1, a2, a3

a1, a2, a3, Á , an, Á

11.1

HISTORICAL ESSAY

Sequences and Series

11.1 Sequences 747



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or by listing terms,

We also sometimes write

Figure 11.1 shows two ways to represent sequences graphically. The first marks thefirst few points from on the real axis. The second method shows thegraph of the function defining the sequence. The function is defined only on integerinputs, and the graph consists of some points in the xy-plane, located at s2, a2d, Á , sn, and, Á .

s1, a1d,

a1, a2, a3, Á , an, Á

5an6 = E2n Fn = 1

q

. .

5dn6 = 51, -1, 1, -1, 1, -1, Á , s -1dn + 1, Á 6 .

5cn6 = e0, 12

, 23

, 34

, 45, Á ,

n - 1n , Á f

5bn6 = e1, -12

, 13

, -14

, Á , s -1dn + 1 1n, Á f

5an6 = E21, 22, 23, Á , 2n, Á F

748 Chapter 11: Infinite Sequences and Series

0

an � �n

1 2

0

Diverges

1 32 4 5

1

3

2

1

Converges to 0

0 1 32 4 5

0

an �

1

0

1

Converges to 0

0

a2 a4 a5 a3 a1

1

1n

n

an

n

an

n

an

a1 a2 a3 a4 a5

a3 a2 a1

an � (�1)n�1 1n

FIGURE 11.1 Sequences can be represented as points on the real line or aspoints in the plane where the horizontal axis n is the index number of theterm and the vertical axis is its value.an

Convergence and Divergence

Sometimes the numbers in a sequence approach a single value as the index n increases.This happens in the sequence

whose terms approach 0 as n gets large, and in the sequence

e0, 12

, 23

, 34

, 45, Á , 1 -

1n, Á f

e1, 12

, 13

, 14

, Á , 1n, Á f



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11.1 Sequences 749

The definition is very similar to the definition of the limit of a function ƒ(x) as x tendsto ( in Section 2.4). We will exploit this connection to calculate limits ofsequences.

EXAMPLE 1 Applying the Definition

Show that

(a) (b)

Solution

(a) Let be given. We must show that there exists an integer N such that for all n,

This implication will hold if or If N is any integer greater thanthe implication will hold for all This proves that

(b) Let be given. We must show that there exists an integer N such that for all n,

Since we can use any positive integer for N and the implication will hold.This proves that for any constant k.limn:q k = k

k - k = 0,

n 7 N Q ƒ k - k ƒ 6 P .

P 7 0

limn:q s1>nd = 0.n 7 N .1>P ,n 7 1>P .s1>nd 6 P

n 7 N Q ` 1n - 0 ` 6 P .

P 7 0

limn: q

k = k sany constant kdlimn: q

1n = 0

limx:q ƒsxdq

whose terms approach 1. On the other hand, sequences like

have terms that get larger than any number as n increases, and sequences like

bounce back and forth between 1 and never converging to a single value. The follow-ing definition captures the meaning of having a sequence converge to a limiting value. Itsays that if we go far enough out in the sequence, by taking the index n to be larger thensome value N, the difference between and the limit of the sequence becomes less thanany preselected number P 7 0.

an

-1,

51, -1, 1, -1, 1, -1, Á , s -1dn + 1, Á 6

E21, 22, 23, Á , 2n, Á F

DEFINITIONS Converges, Diverges, LimitThe sequence converges to the number L if to every positive number therecorresponds an integer N such that for all n,

If no such number L exists, we say that diverges.If converges to L, we write or simply and call

L the limit of the sequence (Figure 11.2).an : L ,limn:q an = L ,5an6

5an6n 7 N Q ƒ an - L ƒ 6 P .

P5an6

aN

(N, aN)

0 1 32 N n

L

L � �

L � � L � �L

L � �

(n, an)

0 a2 a3 a1 an

n

an

FIGURE 11.2 if is ahorizontal asymptote of the sequence ofpoints In this figure, all the after lie within of L.PaN

an’s5sn, and6 .

y = Lan : L

HISTORICAL BIOGRAPHY

Nicole Oresme(ca. 1320–1382)



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EXAMPLE 2 A Divergent Sequence

Show that the sequence diverges.

Solution Suppose the sequence converges to some number L. By choosing inthe definition of the limit, all terms of the sequence with index n larger than some Nmust lie within of L. Since the number 1 appears repeatedly as every other termof the sequence, we must have that the number 1 lies within the distance of L. Itfollows that or equivalently, Likewise, the number appears repeatedly in the sequence with arbitrarily high index. So we must also have that

or equivalently, But the number L cannot lie inboth of the intervals (1 2, 3 2) and because they have no overlap. There-fore, no such limit L exists and so the sequence diverges.

Note that the same argument works for any positive number smaller than 1, notjust 1 2.

The sequence also diverges, but for a different reason. As n increases, itsterms become larger than any fixed number. We describe the behavior of this sequenceby writing

In writing infinity as the limit of a sequence, we are not saying that the differences betweenthe terms and become small as n increases. Nor are we asserting that there is somenumber infinity that the sequence approaches. We are merely using a notation that capturesthe idea that eventually gets and stays larger than any fixed number as n gets large.an

qan

limn: q

2n = q .

{1n}

> P

s -3>2, -1>2d>> -3>2 6 L 6 -1>2.ƒ L - s -1d ƒ 6 1>2,

-11>2 6 L 6 3>2.ƒ L - 1 ƒ 6 1>2,P = 1>2P = 1>2 an

P = 1>251, -1, 1, -1, 1, -1, Á , s -1dn + 1, Á 6


DEFINITION Diverges to InfinityThe sequence diverges to infinity if for every number M there is an integerN such that for all n larger than If this condition holds we write

Similarly if for every number m there is an integer N such that for all wehave then we say diverges to negative infinity and write

limn: q

an = - q or an : - q .

5an6an 6 m ,n 7 N

limn: q

an = q or an : q .

N, an 7 M .5an6

A sequence may diverge without diverging to infinity or negative infinity. We sawthis in Example 2, and the sequences and

are also examples of such divergence.

Calculating Limits of Sequences

If we always had to use the formal definition of the limit of a sequence, calculating with and N’s, then computing limits of sequences would be a formidable task. Fortunately we canderive a few basic examples, and then use these to quickly analyze the limits of many moresequences. We will need to understand how to combine and compare sequences. Since se-quences are functions with domain restricted to the positive integers, it is not too surprisingthat the theorems on limits of functions given in Chapter 2 have versions for sequences.

P’s

51, 0, 2, 0, 3, 0, Á 6 51, -2, 3, -4, 5, -6, 7, -8, Á 6



11.1 Sequences 751

The proof is similar to that of Theorem 1 of Section 2.2, and is omitted.

EXAMPLE 3 Applying Theorem 1

By combining Theorem 1 with the limits of Example 1, we have:

(a)

(b)

(c)

(d)

Be cautious in applying Theorem 1. It does not say, for example, that each of thesequences and have limits if their sum has a limit. For instance,

and both diverge, but their sumclearly converges to 0.

One consequence of Theorem 1 is that every nonzero multiple of a divergent sequencediverges. For suppose, to the contrary, that converges for some number

Then, by taking in the Constant Multiple Rule in Theorem 1, we see that thesequence

converges. Thus, cannot converge unless also converges. If does not con-verge, then does not converge.

The next theorem is the sequence version of the Sandwich Theorem in Section 2.2.You are asked to prove the theorem in Exercise 95.

5can65an65an65can6

e 1c

# can f = 5an6

k = 1>c c Z 0.5can65an65an + bn6 = 50, 0, 0, Á 6 5bn6 = 5-1, -2, -3, Á 65an6 = 51, 2, 3, Á 6 5an + bn65bn65an6

limn: q

4 - 7n6

n6+ 3

= limn: q

s4>n6d - 7

1 + s3>n6d=

0 - 71 + 0

= -7.

limn: q

5n2 = 5 # lim

n: q

1n

# limn: q

1n = 5 # 0 # 0 = 0

limn: q

an - 1n b = lim

n: q

a1 -1n b = lim

n: q

1 - limn: q

1n = 1 - 0 = 1

limn: q

a- 1n b = -1 # lim

n: q

1n = -1 # 0 = 0

THEOREM 1Let and be sequences of real numbers and let A and B be real numbers.The following rules hold if and

1. Sum Rule:

2. Difference Rule:

3. Product Rule:

4. Constant Multiple Rule:

5. Quotient Rule: limn:q an

bn=

AB if B Z 0

limn:q sk # bnd = k # B sAny number kdlimn:q san

# bnd = A # B

limn:q san - bnd = A - B

limn:q san + bnd = A + B

limn:q bn = B .limn:q an = A5bn65an6

Constant Multiple Rule and Example 1a

Difference Ruleand Example 1a

Product Rule

Sum and Quotient Rules

THEOREM 2 The Sandwich Theorem for SequencesLet and be sequences of real numbers. If holdsfor all n beyond some index N, and if then

also.limn:q bn = Llimn:q an = limn:q cn = L ,

an … bn … cn5cn65an6, 5bn6 ,



An immediate consequence of Theorem 2 is that, if and thenbecause We use this fact in the next example.

EXAMPLE 4 Applying the Sandwich Theorem

Since we know that

(a)

(b)

(c)

The application of Theorems 1 and 2 is broadened by a theorem stating that applyinga continuous function to a convergent sequence produces a convergent sequence. We statethe theorem without proof (Exercise 96).

-1n … s -1dn

1n …

1n .because s -1dn

1n : 0

0 …12n …

1n ;because 1

2n : 0 -

1n …

cos nn …

1n ;because cos n

n : 0 1>n : 0,

-cn … bn … cn .bn : 0cn : 0,ƒ bn ƒ … cn


THEOREM 3 The Continuous Function Theorem for SequencesLet be a sequence of real numbers. If and if ƒ is a function that iscontinuous at L and defined at all then ƒsand : ƒsLd .an ,

an : L5an6


Show that

Solution We know that Taking and in Theorem 3gives

EXAMPLE 6 The Sequence

The sequence converges to 0. By taking and inTheorem 3, we see that The sequence convergesto 1 (Figure 11.3).

Using l’Hôpital’s Rule

The next theorem enables us to use l’Hôpital’s Rule to find the limits of some sequences.It formalizes the connection between and limx:q ƒsxd .limn:q an

521>n621>n= ƒs1>nd : ƒsLd = 20

= 1.L = 0an = 1>n, ƒsxd = 2x ,51>n6

521>n61sn + 1d>n : 11 = 1.

L = 1ƒsxd = 1xsn + 1d>n : 1.

2sn + 1d>n : 1.

13

0

1

(1, 2)

y � 2x

1

2

, 21/3

, 21/2

13

12

12

x

y

FIGURE 11.3 As and(Example 6).21>n : 20

n : q , 1>n : 0

THEOREM 4Suppose that ƒ(x) is a function defined for all and that is a sequenceof real numbers such that for Then

limx: q

ƒsxd = L Q limn: q

an = L .

n Ú n0 .an = ƒsnd5an6x Ú n0

Proof Suppose that Then for each positive number there is a num-ber M such that for all x,

x 7 M Q ƒ ƒsxd - L ƒ 6 P .

Plimx:q ƒsxd = L .

4100 AWL/Thomas_ch11p746-847 8/25/04 2:40 PM 52


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11.1 Sequences 753

Let N be an integer greater than M and greater than or equal to Then

EXAMPLE 7 Applying L’Hôpital’s Rule

Show that

Solution The function is defined for all and agrees with the givensequence at positive integers. Therefore, by Theorem 5, will equal

if the latter exists. A single application of l’Hôpital’s Rule shows that

We conclude that

When we use l’Hôpital’s Rule to find the limit of a sequence, we often treat n as acontinuous real variable and differentiate directly with respect to n. This saves us fromhaving to rewrite the formula for as we did in Example 7.

EXAMPLE 8 Applying L’Hôpital’s Rule

Find

Solution By l’Hôpital’s Rule (differentiating with respect to n),

EXAMPLE 9 Applying L’Hôpital’s Rule to Determine Convergence

Does the sequence whose nth term is

converge? If so, find

Solution The limit leads to the indeterminate form We can apply l’Hôpital’s Rule ifwe first change the form to by taking the natural logarithm of

= n ln an + 1n - 1

b .

ln an = ln an + 1n - 1

bn

an :q # 01q .

limn:q an .

an = an + 1n - 1

bn

= q .

limn: q

2n

5n= lim

n: q

2n # ln 2

5

limn: q

2n

5n.

an

limn:q sln nd>n = 0.

limx: q

ln xx = lim

x: q

1>x1

=

01

= 0.

limx:q sln xd>x limn:q sln nd>nx Ú 1sln xd>x

limn: q

ln nn = 0.

n 7 N Q an = ƒsnd and ƒ an - L ƒ = ƒ ƒsnd - L ƒ 6 P .

n0 .



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Then,

Since and is continuous, Theorem 4 tells us that

The sequence converges to

Commonly Occurring Limits

The next theorem gives some limits that arise frequently.

e2 .5an6an = e ln an : e2 .

ƒsxd = exln an : 2

= limn: q

2n2

n2- 1

= 2 .

= limn: q

-2>sn2

- 1d

-1>n2

= limn: q

ln an + 1n - 1

b1>n

limn: q

ln an = limn: q

n ln an + 1n - 1

b


l’Hôpital’s Rule

q # 0

00

THEOREM 5The following six sequences converge to the limits listed below:

1.

2.

3.

4.

5.

6.

In Formulas (3) through (6), x remains fixed as n : q .

limn: q

xn

n!= 0 sany xd

limn: q

a1 +

xn b

n

= ex sany xd

limn: q

xn= 0 s ƒ x ƒ 6 1d

limn: q

x1>n= 1 sx 7 0d

limn: q

2n n = 1

limn: q

ln nn = 0

Proof The first limit was computed in Example 7. The next two can be proved by takinglogarithms and applying Theorem 4 (Exercises 93 and 94). The remaining proofs are givenin Appendix 3.


(a) Formula 1

(b) Formula 2

(c) Formula 3 with and Formula 2x = 32n 3n = 31>nsn1/nd : 1 # 1 = 1

2n n2= n2>n

= sn1/nd2 : s1d2= 1

ln sn2dn =

2 ln nn : 2 # 0 = 0

Factorial NotationThe notation n! (“n factorial”) means the product of the integersfrom 1 to n. Notice that

Thus,and

Wedefine 0! to be 1. Factorials grow evenfaster than exponentials, as the tablesuggests.

5! = 1 # 2 # 3 # 4 # 5 = 5 # 4! = 120.4! = 1 # 2 # 3 # 4 = 24sn + 1d! = sn + 1d # n! .

1 # 2 # 3 Á n

n (rounded) n!

1 3 1

5 148 120

10 22,026 3,628,800

20 2.4 * 10184.9 * 108

en



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11.1 Sequences 755

(d) Formula 4 with

(e) Formula 5 with

(f) Formula 6 with

Recursive Definitions

So far, we have calculated each directly from the value of n. But sequences are oftendefined recursively by giving

1. The value(s) of the initial term or terms, and

2. A rule, called a recursion formula, for calculating any later term from terms that pre-cede it.

EXAMPLE 11 Sequences Constructed Recursively

(a) The statements and define the sequence ofpositive integers. With we have andso on.

(b) The statements and define the sequence of factorials. With we have

and so on.

(c) The statements and define the sequenceof Fibonacci numbers. With and we have

and so on.

(d) As we can see by applying Newton’s method, the statements anddefine a sequence that converges to a

solution of the equation

Bounded Nondecreasing Sequences

The terms of a general sequence can bounce around, sometimes getting larger, sometimessmaller. An important special kind of sequence is one for which each term is at least aslarge as its predecessor.

sin x - x2= 0.

xn + 1 = xn - [ssin xn - xn2d>scos xn - 2xnd]

x0 = 1

a3 = 1 + 1 = 2, a4 = 2 + 1 = 3, a5 = 3 + 2 = 5,a2 = 1,a1 = 11, 1, 2, 3, 5, Á

an + 1 = an + an - 1a1 = 1, a2 = 1,

4 # a3 = 24,a2 = 2 # a1 = 2, a3 = 3 # a2 = 6, a4 =a1 = 1,

1, 2, 6, 24, Á , n!, Áan = n # an - 1a1 = 1

a2 = a1 + 1 = 2, a3 = a2 + 1 = 3,a1 = 1,1, 2, 3, Á , n, Áan = an - 1 + 1a1 = 1

an

x = 100100n

n!: 0

x = -2an - 2n bn

= a1 +-2n b

n

: e-2

x = -

12

a- 12bn

: 0

DEFINITION Nondecreasing SequenceA sequence with the property that for all n is called anondecreasing sequence.

an … an + 15an6

EXAMPLE 12 Nondecreasing Sequences

(a) The sequence of natural numbers

(b) The sequence

(c) The constant sequence 53612

, 23

, 34

, Á , n

n + 1, Á

1, 2, 3, Á , n, Á



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There are two kinds of nondecreasing sequences—those whose terms increase beyond anyfinite bound and those whose terms do not.


DEFINITIONS Bounded, Upper Bound, Least Upper BoundA sequence is bounded from above if there exists a number M such that

for all n. The number M is an upper bound for If M is an upperbound for but no number less than M is an upper bound for then M isthe least upper bound for 5an6 .

5an6 ,5an65an6 .an … M

5an6

EXAMPLE 13 Applying the Definition for Boundedness

(a) The sequence has no upper bound.

(b) The sequence is bounded above by

No number less than 1 is an upper bound for the sequence, so 1 is the least upperbound (Exercise 113).

A nondecreasing sequence that is bounded from above always has a least upperbound. This is the completeness property of the real numbers, discussed in Appendix 4.We will prove that if L is the least upper bound then the sequence converges to L.

Suppose we plot the points in the xy-plane. If M is an up-per bound of the sequence, all these points will lie on or below the line (Figure 11.4).The line is the lowest such line. None of the points lies above but somedo lie above any lower line if is a positive number. The sequence converges toL because

(a) for all values of n and

(b) given any there exists at least one integer N for which

The fact that is nondecreasing tells us further that

Thus, all the numbers beyond the Nth number lie within of L. This is precisely thecondition for L to be the limit of the sequence

The facts for nondecreasing sequences are summarized in the following theorem. Asimilar result holds for nonincreasing sequences (Exercise 107).

{an}.Pan

an Ú aN 7 L - P for all n Ú N .

5an6aN 7 L - P .P 7 0,

an … L

Py = L - P ,y = L ,sn, andy = L

y = Ms1, a1d, s2, a2d, Á , sn, and, Á

M = 1.12

, 23

, 34

, Á , n

n + 1, Á

1, 2, 3, Á , n, Á

0 1 2 3 4

L

M

5

y � L

(8, a8)

6 7 8

y � M

(5, a5)

(1, a1)

x

y

FIGURE 11.4 If the terms of anondecreasing sequence have an upperbound M, they have a limit L … M .

THEOREM 6 The Nondecreasing Sequence TheoremA nondecreasing sequence of real numbers converges if and only if it is boundedfrom above. If a nondecreasing sequence converges, it converges to its leastupper bound.

Theorem 6 implies that a nondecreasing sequence converges when it is bounded fromabove. It diverges to infinity if it is not bounded from above.



11.1 Sequences 757

EXERCISES 11.1

Finding Terms of a SequenceEach of Exercises 1–6 gives a formula for the nth term of a se-quence Find the values of and

1. 2.

3. 4.

5. 6.

Each of Exercises 7–12 gives the first term or two of a sequence alongwith a recursion formula for the remaining terms. Write out the firstten terms of the sequence.

7.

8.

9.

10.

11.

12.

Finding a Sequence’s FormulaIn Exercises 13–22, find a formula for the nth term of the sequence.

13. The sequence

14. The sequence

15. The sequence

16. The sequence

17. The sequence

18. The sequence

19. The sequence

20. The sequence

21. The sequence

22. The sequence

Finding LimitsWhich of the sequences in Exercises 23–84 converge, and whichdiverge? Find the limit of each convergent sequence.

5an6

0, 1, 1, 2, 2, 3, 3, 4, Á

1, 0, 1, 0, 1, Á

2, 6, 10, 14, 18, Á

1, 5, 9, 13, 17, Á

-3, -2, -1, 0, 1, Á

0, 3, 8, 15, 24, Á

1, -14

, 19

, -1

16,

125

, Á

1, -4, 9, -16, 25, Á

-1, 1, -1, 1, -1, Á

1, -1, 1, -1, 1, Á

a1 = 2, a2 = -1, an + 2 = an + 1>an

a1 = a2 = 1, an + 2 = an + 1 + an

a1 = -2, an + 1 = nan>sn + 1da1 = 2, an + 1 = s -1dn + 1an>2a1 = 1, an + 1 = an>sn + 1da1 = 1, an + 1 = an + s1>2nd

an =

2n- 1

2nan =

2n

2n + 1

an = 2 + s -1dnan =

s -1dn + 1

2n - 1

an =

1n!

an =

1 - n

n2

a4 .a1, a2, a3 ,5an6 .an

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

57. 58.

59. (Hint: Compare with 1 n.)>an =

n!nn

an = 2n 32n + 1an = 2n 4nn

an = ln n - ln sn + 1dan =

ln n

n1>n

an = sn + 4d1>sn + 4dan = a3n b1>n

an = 2n n2an = 2n 10n

an = a1 -

1n b

n

an = a1 +

7n b

n

an = s0.03d1>nan = 81>n

an =

ln nln 2n

an =

ln sn + 1d2n

an =

3n

n3an =

n2n

an =

sin2 n2nan =

sin nn

an = np cos snpdan = sin ap2

+

1n b

an =

1s0.9dnan = A 2n

n + 1

an = a- 12bn

an =

s -1dn + 1

2n - 1

an = a2 -

12n b a3 +

12n ban = an + 1

2nb a1 -

1n b

an = s -1dn a1 -

1n ban = 1 + s -1dn

an =

1 - n3

70 - 4n2an =

n2- 2n + 1n - 1

an =

n + 3n2

+ 5n + 6an =

1 - 5n4

n4+ 8n3

an =

2n + 1

1 - 32nan =

1 - 2n1 + 2n

an =

n + s -1dn

nan = 2 + s0.1dn

Reciprocals of squaresof the positive integers,with alternating signs

1’s with alternating signs

1’s with alternating signs

Squares of the positiveintegers; withalternating signs

Squares of the positiveintegers diminished by 1

Integers beginning with-3

Every other odd positiveinteger

Every other even positiveinteger

Alternating 1’s and 0’s

Each positive integerrepeated



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60. 61.

62. 63.

64. 65.

66. 67.

68. 69.

70. 71.

72. 73.

74. 75.

76. 77.

78. 79.

80. 81.

82.

83. 84.

Theory and Examples85. The first term of a sequence is Each succeeding term is

the sum of all those that come before it:

Write out enough early terms of the sequence to deduce a generalformula for that holds for

86. A sequence of rational numbers is described as follows:

.

Here the numerators form one sequence, the denominators form asecond sequence, and their ratios form a third sequence. Let and be, respectively, the numerator and the denominator of thenth fraction

a. Verify that and, moregenerally, that if or then

respectively.

sa + 2bd2- 2sa + bd2

= +1 or -1,

+1,a2- 2b2

= -1x1

2- 2y1

2= -1, x2

2- 2y2

2= +1

rn = xn>yn .yn

xn

11

, 32

, 75

, 1712

, Á , ab

, a + 2ba + b

, Á

n Ú 2.xn

xn + 1 = x1 + x2 +Á

+ xn .

x1 = 1.

an = Ln

1 1xp dx, p 7 1an =

1nL

n

1 1x dx

an =

12n2- 1 - 2n2

+ n

an = n - 2n2- nan =

sln nd52n

an =

sln nd200

nan = 2n n2+ n

an = a13bn

+

122nan =

12n tan-1 n

an = tan-1 nan = n a1 - cos 1n b

an =

n2

2n - 1 sin

1nan = sinh sln nd

an = tanh nan =

s10>11dn

s9/10dn+ s11/12dn

an =

3n # 6n

2-n # n!an = a1 -

1n2 b

n

an = a xn

2n + 1b1>n

, x 7 0an = a nn + 1

bn

an = a3n + 13n - 1

bn

an = ln a1 +

1n b

n

an = a1n b1>sln nd

an =

n!2n # 3n

an =

n!106n

an =

s -4dn

n!b. The fractions approach a limit as n increases.

What is that limit? (Hint: Use part (a) to show thatand that is not less than n.)

87. Newton’s method The following sequences come from the re-cursion formula for Newton’s method,

Do the sequences converge? If so, to what value? In each case,begin by identifying the function ƒ that generates the sequence.

a.

b.

c.

88. a. Suppose that ƒ(x) is differentiable for all x in [0, 1] and thatDefine the sequence by the rule

Show that

Use the result in part (a) to find the limits of the followingsequences

b. c.

d.

89. Pythagorean triples A triple of positive integers a, b, and c iscalled a Pythagorean triple if Let a be an oddpositive integer and let

be, respectively, the integer floor and ceiling for

a. Show that (Hint: Let and expressb and c in terms of n.)

a = 2n + 1a2+ b2

= c2 .

a

a2

2

a2

2

�

a2>2.

b = j a2

2k and c = l a2

2m

a2+ b2

= c2 .

an = n ln a1 +

2n b

an = nse1>n- 1dan = n tan-1

1n

5an6 .

lim n:q an = ƒ¿s0d .nƒs1>nd .an =5an6ƒs0d = 0.

x0 = 1, xn + 1 = xn - 1

x0 = 1, xn + 1 = xn -

tan xn - 1

sec2 xn

x0 = 1, xn + 1 = xn -

xn2

- 22xn

=

xn

2+

1xn

xn + 1 = xn -

ƒsxndƒ¿sxnd

.

ynrn2

- 2 = ;s1>ynd2

rn = xn>yn




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11.1 Sequences 759

b. By direct calculation, or by appealing to the figure here, find

90. The nth root of n!

a. Show that and hence, using Stirling’sapproximation (Chapter 8, Additional Exercise 50a), that

b. Test the approximation in part (a) for asfar as your calculator will allow.

91. a. Assuming that if c is any positive con-stant, show that

if c is any positive constant.

b. Prove that if c is any positive constant.(Hint: If and how large should N be toensure that if )

92. The zipper theorem Prove the “zipper theorem” for se-quences: If and both converge to L, then the sequence

converges to L.

93. Prove that

94. Prove that

95. Prove Theorem 2. 96. Prove Theorem 3.

In Exercises 97–100, determine if the sequence is nondecreasing andif it is bounded from above.

97. 98.

99. 100.

Which of the sequences in Exercises 101–106 converge, and which di-verge? Give reasons for your answers.

101. 102.

103. 104.

105. an = ss -1dn+ 1d an + 1

n ban =

2n- 1

3nan =

2n- 1

2n

an = n -

1nan = 1 -

1n

an = 2 -

2n -

12nan =

2n3n

n!

an =

s2n + 3d!sn + 1d!

an =

3n + 1n + 1

limn:q x1>n= 1, sx 7 0d .

limn:q2n n = 1.

a1, b1, a2 , b2 , Á , an , bn , Á

5bn65an6n 7 N?ƒ 1>nc

- 0 ƒ 6 P

c = 0.04 ,P = 0.001limn:q s1>ncd = 0

limn: q

ln nnc = 0

limn:q s1>ncd = 0

n = 40, 50, 60, Á ,

2n n! L

ne for large values of n .

limn:q s2npd1>s2nd= 1

lima: q

j a2

2k

l a2

2m.

106. The first term of a sequence is The next terms areor cos (2), whichever is larger; and or cos (3),

whichever is larger (farther to the right). In general,

107. Nonincreasing sequences A sequence of numbers inwhich for every n is called a nonincreasing sequence.A sequence is bounded from below if there is a number Mwith for every n. Such a number M is called a lowerbound for the sequence. Deduce from Theorem 6 that a nonin-creasing sequence that is bounded from below converges and thata nonincreasing sequence that is not bounded from below di-verges.

(Continuation of Exercise 107.) Using the conclusion of Exercise 107,determine which of the sequences in Exercises 108–112 converge andwhich diverge.

108. 109.

110. 111.

112.

113. The sequence has a least upper bound of 1Show that if M is a number less than 1, then the terms of

eventually exceed M. That is, if there is aninteger N such that whenever Since

for every n, this proves that 1 is a least upperbound for

114. Uniqueness of least upper bounds Show that if and are least upper bounds for the sequence then That is, a sequence cannot have two different least upper bounds.

115. Is it true that a sequence of positive numbers must con-verge if it is bounded from above? Give reasons for your answer.

116. Prove that if is a convergent sequence, then to every pos-itive number there corresponds an integer N such that for allm and n,

117. Uniqueness of limits Prove that limits of sequences areunique. That is, show that if and are numbers such that

and then

118. Limits and subsequences If the terms of one sequence ap-pear in another sequence in their given order, we call the firstsequence a subsequence of the second. Prove that if two sub-sequences of a sequence have different limits then diverges.

119. For a sequence the terms of even index are denoted by and the terms of odd index by Prove that if and

then

120. Prove that a sequence converges to 0 if and only if the se-quence of absolute values converges to 0.5ƒ an ƒ6

5an6an : L .a2k + 1 : L ,

a2k : La2k + 1 .a2k5an6

5an6L1 Z L2 ,5an6

L1 = L2 .an : L2 ,an : L1

L2L1

m 7 N and n 7 N Q ƒ am - an ƒ 6 P .

P

5an65an6

M1 = M2 .5an6 ,M2M1

5n>sn + 1d6 .n>sn + 1d 6 1

n 7 N .n>sn + 1d 7 MM 6 15n>sn + 1d6

5n>sn + 1d6a1 = 1, an + 1 = 2an - 3

an =

4n + 1+ 3n

4nan =

1 - 4n

2n

an =

1 + 22n2nan =

n + 1n

M … an

5an6an Ú an + 1

5an6xn + 1 = max 5xn , cos sn + 1d6 .

x3 = x2x2 = x1

x1 = cos s1d .

T



Calculator Explorations of LimitsIn Exercises 121–124, experiment with a calculator to find a value ofN that will make the inequality hold for all Assuming that theinequality is the one from the formal definition of the limit of a se-quence, what sequence is being considered in each case and what is itslimit?

121. 122.

123. 124.

125. Sequences generated by Newton’s method Newton’smethod, applied to a differentiable function ƒ(x), begins with astarting value and constructs from it a sequence of numbers

that under favorable circumstances converges to a zero ofƒ. The recursion formula for the sequence is

a. Show that the recursion formula for can be written as

b. Starting with and calculate successive termsof the sequence until the display begins to repeat. What num-ber is being approximated? Explain.

126. (Continuation of Exercise 125.) Repeat part (b) of Exercise 125with in place of

127. A recursive definition of If you start with anddefine the subsequent terms of by the rule

you generate a sequence that convergesrapidly to a. Try it. b. Use the accompanying figure to ex-plain why the convergence is so rapid.

128. According to a front-page article in the December 15, 1992, is-sue of the Wall Street Journal, Ford Motor Company used about

hours of labor to produce stampings for the average vehicle,down from an estimated 15 hours in 1980. The Japanese neededonly about hours.

Ford’s improvement since 1980 represents an average de-crease of 6% per year. If that rate continues, then n years from1992 Ford will use about

hours of labor to produce stampings for the average vehicle. As-suming that the Japanese continue to spend hours per vehicle,3 12

Sn = 7.25s0.94dn

3 12

7 14

10

cos xn � 11

xn � 1

xn � 1x

y

p>2.xn = xn - 1 + cos xn - 1 ,

5xn6x1 = 1P/2

a = 3.a = 2

a = 3,x0 = 1

xn + 1 = sxn + a>xnd>2.ƒsxd = x2

- a, a 7 0,

xn + 1 = xn -

ƒsxndƒ¿sxnd

.

5xn6x0

2n>n! 6 10-7s0.9dn6 10-3

ƒ2n n - 1 ƒ 6 10-3ƒ2n 0.5 - 1 ƒ 6 10-3

n 7 N .

how many more years will it take Ford to catch up? Find out twoways:

a. Find the first term of the sequence that is less than orequal to 3.5.

b. Graph and use Trace to find where thegraph crosses the line

COMPUTER EXPLORATIONS

Use a CAS to perform the following steps for the sequences in Exer-cises 129–140.

a. Calculate and then plot the first 25 terms of the sequence. Doesthe sequence appear to be bounded from above or below? Does itappear to converge or diverge? If it does converge, what is thelimit L?

b. If the sequence converges, find an integer N such thatfor How far in the sequence do you

have to get for the terms to lie within 0.0001 of L?

129. 130.

131.

132.

133. 134.

135. 136.

137. 138.

139. 140.

141. Compound interest, deposits, and withdrawals If you investan amount of money at a fixed annual interest rate r com-pounded m times per year, and if the constant amount b is addedto the account at the end of each compounding period (or takenfrom the account if ), then the amount you have after

compounding periods is

(1)

a. If and calculateand plot the first 100 points How much money is inyour account at the end of 5 years? Does converge? Is

bounded?

b. Repeat part (a) with and

c. If you invest 5000 dollars in a certificate of deposit (CD) thatpays 4.5% annually, compounded quarterly, and you make nofurther investments in the CD, approximately how manyyears will it take before you have 20,000 dollars? What if theCD earns 6.25%?

b = -50.A0 = 5000, r = 0.0589, m = 12,

5An65An6

sn, And .b = 50,A0 = 1000, r = 0.02015, m = 12,

An + 1 = a1 +

rm bAn + b .

n + 1b 6 0

A0

an =

n41

19nan =

8n

n!

an = 1234561>nan = s0.9999dn

an =

ln nnan =

sin nn

an = n sin 1nan = sin n

a1 = 1, an + 1 = an + s -2dn

a1 = 1, an + 1 = an +

15n

an = a1 +

0.5n b

n

an = 2n n

n Ú N .ƒ an - L ƒ … 0.01

y = 3.5 .ƒsxd = 7.25s0.94dx

5Sn6


T

T



761

d. It can be shown that for any the sequence defined re-cursively by Equation (1) satisfies the relation

(2)

For the values of the constants r, m, and b given in part(a), validate this assertion by comparing the values of thefirst 50 terms of both sequences. Then show by direct substi-tution that the terms in Equation (2) satisfy the recursion for-mula in Equation (1).

142. Logistic difference equation The recursive relation

is called the logistic difference equation, and when the initialvalue is given the equation defines the logistic sequence

Throughout this exercise we choose in the intervalsay

a. Choose Calculate and plot the points for thefirst 100 terms in the sequence. Does it appear to converge?What do you guess is the limit? Does the limit seem to de-pend on your choice of

b. Choose several values of r in the interval and re-peat the procedures in part (a). Be sure to choose some pointsnear the endpoints of the interval. Describe the behavior ofthe sequences you observe in your plots.

c. Now examine the behavior of the sequence for values of rnear the endpoints of the interval The transi-tion value is called a bifurcation value and the newbehavior of the sequence in the interval is called anattracting 2-cycle. Explain why this reasonably describes thebehavior.

r = 33 6 r 6 3.45 .

1 6 r 6 3

a0 ?

sn, andr = 3>4.

a0 = 0.3 .0 6 a0 6 1,a05an6 .

a0

an + 1 = rans1 - and

A0 ,

Ak = a1 +

rm b

k

aA0 +

mbr b -

mbr .

k Ú 0, d. Next explore the behavior for r values near the endpointsof each of the intervals and

Plot the first 200 terms of the sequences.Describe in your own words the behavior observed in yourplots for each interval. Among how many values does the se-quence appear to oscillate for each interval? The values

and (rounded to two decimal places) arealso called bifurcation values because the behavior of the se-quence changes as r crosses over those values.

e. The situation gets even more interesting. There is actually anincreasing sequence of bifurcation values

such that for thelogistic sequence eventually oscillates steadily among

values, called an attracting Moreover, the bifur-cation sequence is bounded above by 3.57 (so it con-verges). If you choose a value of you will observe a

of some sort. Choose and plot 300points.

f. Let us see what happens when Choose and calculate and plot the first 300 terms of Observehow the terms wander around in an unpredictable, chaoticfashion. You cannot predict the value of from previousvalues of the sequence.

g. For choose two starting values of that are closetogether, say, and Calculate and plotthe first 300 values of the sequences determined by eachstarting value. Compare the behaviors observed in yourplots. How far out do you go before the correspondingterms of your two sequences appear to depart from eachother? Repeat the exploration for Can you seehow the plots look different depending on your choice of

We say that the logistic sequence is sensitive to the ini-tial condition a0 .a0 ?

r = 3.75 .

a0 = 0.301 .a0 = 0.3a0r = 3.65

an + 1

5an6 .r = 3.65r 7 3.57 .

r = 3.56952n-cycler 6 3.57

5cn62n-cycle.2n

5an6cn 6 r 6 cn + 16

Á6 cn 6 cn + 1

Á

3 6 3.45 6 3.54

r = 3.54r = 3.45

3.54 6 r 6 3.55 .3.45 6 r 6 3.54


11.1 Sequences


11.2 Infinite Series 761

Infinite Series

An infinite series is the sum of an infinite sequence of numbers

The goal of this section is to understand the meaning of such an infinite sum and to de-velop methods to calculate it. Since there are infinitely many terms to add in an infinite se-ries, we cannot just keep adding to see what comes out. Instead we look at what we get bysumming the first n terms of the sequence and stopping. The sum of the first n terms

sn = a1 + a2 + a3 +Á

+ an

a1 + a2 + a3 +Á

+ an +Á

11.2



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is an ordinary finite sum and can be calculated by normal addition. It is called the nth par-tial sum. As n gets larger, we expect the partial sums to get closer and closer to a limitingvalue in the same sense that the terms of a sequence approach a limit, as discussed inSection 11.1.

For example, to assign meaning to an expression like

We add the terms one at a time from the beginning and look for a pattern in how these par-tial sums grow.

Suggestiveexpression for

Partial sum partial sum Value

First: 1

Second:

Third:

nth:

Indeed there is a pattern. The partial sums form a sequence whose nth term is

This sequence of partial sums converges to 2 because We say

Is the sum of any finite number of terms in this series equal to 2? No. Can we actually addan infinite number of terms one by one? No. But we can still define their sum by definingit to be the limit of the sequence of partial sums as in this case 2 (Figure 11.5).Our knowledge of sequences and limits enables us to break away from the confines offinite sums.

n : q ,

“the sum of the infinite series 1 +12

+14

+Á

+1

2n - 1 +Á is 2.”

limn:q s1>2nd = 0.

sn = 2 -1

2n - 1 .

2n- 1

2n - 12 -1

2n - 1 sn = 1 +12

+14

+Á

+1

2n - 1

ooo o

74

2 -14

s3 = 1 +12

+14

32

2 -12

s2 = 1 +12

2 - 1 s1 = 1

1 +12

+14

+18

+116

+Á

0

1

1 21/2 1/8

1/4

FIGURE 11.5 As the lengths are added one by one, the sumapproaches 2.

1, 1�2, 1�4, 1�8, Á





When we begin to study a given series we might not knowwhether it converges or diverges. In either case, it is convenient to use sigma notation towrite the series as

Geometric Series

Geometric series are series of the form

in which a and r are fixed real numbers and The series can also be written asThe ratio r can be positive, as in

or negative, as in

If the nth partial sum of the geometric series is

sn = a + as1d + as1d2+

Á+ as1dn - 1

= na ,

r = 1,

1 -13

+19

-Á

+ a- 13bn - 1

+Á .

1 +12

+14

+Á

+ a12bn - 1

+Á ,

gq

n=0 arn .a Z 0.

a + ar + ar2+

Á+ arn - 1

+Á

= aq

n = 1 arn - 1

aq

n = 1 an, a

q

k = 1 ak, or a an

a1 + a2 +Á

+ an +Á ,

DEFINITIONS Infinite Series, nth Term, Partial Sum, Converges, SumGiven a sequence of numbers an expression of the form

is an infinite series. The number is the nth term of the series. The sequencedefined by

is the sequence of partial sums of the series, the number being the nth partialsum. If the sequence of partial sums converges to a limit L, we say that the seriesconverges and that its sum is L. In this case, we also write

If the sequence of partial sums of the series does not converge, we say that theseries diverges.

a1 + a2 +Á

+ an +Á

= aq

n = 1 an = L .

sn

o

sn = a1 + a2 +Á

+ an = an

k = 1 ak

o

s2 = a1 + a2

s1 = a1

5sn6an

a1 + a2 + a3 +Á

+ an +Á

5an6 ,


Blaise Pascal(1623–1662)

A useful shorthandwhen summationfrom 1 to isunderstood

q



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If the geometric series convergesto

If the series diverges.ƒ r ƒ Ú 1,

aq

n = 1 arn - 1

=

a1 - r

, ƒ r ƒ 6 1.

a>s1 - rd :a + ar + ar2

+Á

+ arn - 1+

Áƒ r ƒ 6 1,

and the series diverges because depending on the sign of a. If the series diverges because the nth partial sums alternate between a and 0. If wecan determine the convergence or divergence of the series in the following way:

If then as (as in Section 11.1) and If then and the series diverges.ƒ rn

ƒ : q

ƒ r ƒ 7 1,sn : a>s1 - rd .n : qrn : 0ƒ r ƒ 6 1,

sn =

as1 - rnd1 - r

, sr Z 1d .

sns1 - rd = as1 - rnd sn - rsn = a - arn

rsn = ar + ar2+

Á+ arn - 1

+ arn

sn = a + ar + ar2+

Á+ arn - 1

ƒ r ƒ Z 1,r = -1,limn:q sn = ; q ,


Subtract from Most ofthe terms on the right cancel.

sn .rsn

Factor.

We can solve for sn if r Z 1 .

Multiply by r.sn

We have determined when a geometric series converges or diverges, and to whatvalue. Often we can determine that a series converges without knowing the value to whichit converges, as we will see in the next several sections. The formula for the sumof a geometric series applies only when the summation index begins with in the ex-pression (or with the index if we write the series as ).

EXAMPLE 1 Index Starts with

The geometric series with and is

EXAMPLE 2 Index Starts with

The series

is a geometric series with and It converges to

EXAMPLE 3 A Bouncing Ball

You drop a ball from a meters above a flat surface. Each time the ball hits the surface afterfalling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find thetotal distance the ball travels up and down (Figure 11.6).

a1 - r

=

51 + s1>4d

= 4.

r = -1>4.a = 5

aq

n = 0 s -1dn5

4n = 5 -

54

+

516

-

564

+Á

n = 0

19

+127

+181

+Á

= aq

n = 1 19

a13bn - 1

=

1>91 - s1>3d

=16

.

r = 1>3a = 1>9n = 1

gq

n=0 arnn = 0gq

n=1 arn - 1n = 1

a>s1 - rd



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Solution The total distance is

If and for instance, the distance is

EXAMPLE 4 Repeating Decimals

Express the repeating decimal as the ratio of two integers.

Solution

Unfortunately, formulas like the one for the sum of a convergent geometric series are rareand we usually have to settle for an estimate of a series’ sum (more about this later). Thenext example, however, is another case in which we can find the sum exactly.

EXAMPLE 5 A Nongeometric but Telescoping Series

Find the sum of the series

Solution We look for a pattern in the sequence of partial sums that might lead to a for-mula for The key observation is the partial fraction decomposition

so

and

Removing parentheses and canceling adjacent terms of opposite sign collapses the sum to

sk = 1 -1

k + 1.

sk = a11

-12b + a1

2-

13b + a1

3-

14b +

Á+ a1

k-

1k + 1

b .

ak

n = 1

1nsn + 1d

= ak

n = 1 a1n -

1n + 1

b

1nsn + 1d

=1n -

1n + 1

,

sk .

aq

n = 1

1nsn + 1d

.

= 5 +

23100

a 10.99b = 5 +

2399

=

51899

1>s1 - 0.01d('''''''')''''''''*

= 5 +

23100

a1 +1

100+ a 1

100b2

+Á b

5.232323 Á = 5 +

23100

+

23s100d2 +

23s100d3 +

Á

5.232323 Á

s = 6 1 + s2>3d1 - s2>3d

= 6 a5>31>3 b = 30 m.

r = 2>3,a = 6 m

This sum is 2ar>s1 - rd.(''''''')'''''''*

s = a + 2ar + 2ar2+ 2ar3

+Á

= a +

2ar1 - r

= a 1 + r1 - r

.

r = 1>100a = 1 ,

ar

ar2

ar3

(a)

a

FIGURE 11.6 (a) Example 3 shows howto use a geometric series to calculate thetotal vertical distance traveled by abouncing ball if the height of each reboundis reduced by the factor r. (b) Astroboscopic photo of a bouncing ball.

(b)



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We now see that as The series converges, and its sum is 1:

Divergent Series

One reason that a series may fail to converge is that its terms don’t become small.

EXAMPLE 6 Partial Sums Outgrow Any Number

(a) The series

diverges because the partial sums grow beyond every number L. After the par-tial sum is greater than

(b) The series

diverges because the partial sums eventually outgrow every preassigned number. Eachterm is greater than 1, so the sum of n terms is greater than n.

The nth-Term Test for Divergence

Observe that must equal zero if the series converges. To see why, let Srepresent the series’ sum and the nth partial sum. When n islarge, both and are close to S, so their difference, is close to zero. More formally,

This establishes the following theorem.

an = sn - sn - 1 : S - S = 0.

an ,sn - 1sn

sn = a1 + a2 +Á

+ an

gq

n=1 anlimn:q an

aq

n = 1 n + 1

n =21

+

32

+43

+Á

+

n + 1n +

Á

n2 .sn = 1 + 4 + 9 +Á

+ n2n = 1,

aq

n = 1 n2

= 1 + 4 + 9 +Á

+ n2+

Á

aq

n = 1

1nsn + 1d

= 1.

k : q .sk : 1


Difference Rule forsequences

THEOREM 7

If converges, then an : 0.aq

n = 1 an

Theorem 7 leads to a test for detecting the kind of divergence that occurred in Example 6.

CautionTheorem 7 does not say that converges if It is possible for aseries to diverge when an : 0.

an : 0.gq

n=1 an

The nth-Term Test for Divergence

diverges if fails to exist or is different from zero.limn: q

anaq

n = 1 an




EXAMPLE 7 Applying the nth-Term Test

(a) diverges because

(b) diverges because

(c) diverges because does not exist

(d) diverges because

EXAMPLE 8 but the Series Diverges

The series

2 terms 4 terms

diverges because the terms are grouped into clusters that add to 1, so the partial sumsincrease without bound. However, the terms of the series form a sequence that con-verges to 0. Example 1 of Section 11.3 shows that the harmonic series also behaves inthis manner.

Combining Series

Whenever we have two convergent series, we can add them term by term, subtract themterm by term, or multiply them by constants to make new convergent series.

2n terms(''''')'''''*('''')''''*(')'*

1 +12

+12

+14

+14

+14

+14

+Á

+12n +

12n +

Á+

12n +

Á

an : 0

limn:q -n

2n + 5= -

12

Z 0.aq

n = 1

-n2n + 5

limn:qs -1dn + 1aq

n = 1 s -1dn + 1

n + 1n : 1a

q

n = 1 n + 1

n

n2 : qaq

n = 1 n2

THEOREM 8If and are convergent series, then

1. Sum Rule:

2. Difference Rule:

3. Constant Multiple Rule: gkan = kgan = kA sAny number kd .

gsan - bnd = gan - gbn = A - B

gsan + bnd = gan + gbn = A + B

gbn = Bgan = A

Proof The three rules for series follow from the analogous rules for sequences inTheorem 1, Section 11.1. To prove the Sum Rule for series, let

Then the partial sums of are

= An + Bn .

= sa1 +Á

+ and + sb1 +Á

+ bnd sn = sa1 + b1d + sa2 + b2d +

Á+ san + bnd

gsan + bnd

An = a1 + a2 +Á

+ an, Bn = b1 + b2 +Á

+ bn .



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Since and we have by the Sum Rule for sequences. Theproof of the Difference Rule is similar.

To prove the Constant Multiple Rule for series, observe that the partial sums of form the sequence

which converges to kA by the Constant Multiple Rule for sequences.As corollaries of Theorem 8, we have

1. Every nonzero constant multiple of a divergent series diverges.

2. If converges and diverges, then and both diverge.

We omit the proofs.

CAUTION Remember that can converge when and both diverge.For example, and diverge,whereas converges to 0.

EXAMPLE 9 Find the sums of the following series.

(a)

(b)

Adding or Deleting Terms

We can add a finite number of terms to a series or delete a finite number of terms withoutaltering the series’ convergence or divergence, although in the case of convergence this willusually change the sum. If converges, then converges for any and

Conversely, if converges for any then converges. Thus,

aq

n = 1 15n =

15 +

125

+1

125+ a

q

n = 4 15n

gq

n=1 ank 7 1,gq

n=k an

aq

n = 1 an = a1 + a2 +

Á+ ak - 1 + a

q

n = k an .

k 7 1gq

n=k angq

n=1 an

= 8

= 4 a 11 - s1>2d

b

aq

n = 0 42n = 4a

q

n = 0 12n

=45

= 2 -

65

=1

1 - s1>2d-

11 - s1>6d

= aq

n = 1

12n - 1 - a

q

n = 1

16n - 1

aq

n = 1 3n - 1

- 16n - 1 = a

q

n = 1 a 1

2n - 1 -1

6n - 1 b

gsan + bnd = 0 + 0 + 0 +Á

gbn = s -1d + s -1d + s -1d +Ágan = 1 + 1 + 1 +

Á

gbngangsan + bnd

gsan - bndgsan + bndgbngan

sn = ka1 + ka2 +Á

+ kan = ksa1 + a2 +Á

+ and = kAn ,

gkan

sn : A + BBn : B ,An : A


Difference Rule

Geometric series with a = 1 and r = 1>2, 1>6

Constant Multiple Rule

Geometric series with a = 1, r = 1>2



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and

Reindexing

As long as we preserve the order of its terms, we can reindex any series without altering itsconvergence. To raise the starting value of the index h units, replace the n in the formulafor by

To lower the starting value of the index h units, replace the n in the formula for by

It works like a horizontal shift. We saw this in starting a geometric series with the indexinstead of the index but we can use any other starting index value as well.

We usually give preference to indexings that lead to simple expressions.

EXAMPLE 10 Reindexing a Geometric Series

We can write the geometric series

as

The partial sums remain the same no matter what indexing we choose.

aq

n = 0 12n, a

q

n = 5

12n - 5, or even a

q

n = -4

12n + 4 .

aq

n = 1

12n - 1 = 1 +

12

+14

+Á

n = 1,n = 0

aq

n = 1 an = a

q

n = 1 - h an + h = a1 + a2 + a3 +

Á .

n + h :an

aq

n = 1 an = a

q

n = 1 + h an - h = a1 + a2 + a3 +

Á .

n - h :an

aq

n = 4 15n = aa

q

n = 1 15n b -

15 -

125

-1

125.


Richard Dedekind(1831–1916)



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EXERCISES 11.2

Finding nth Partial SumsIn Exercises 1–6, find a formula for the nth partial sum of each seriesand use it to find the series’ sum if the series converges.

1.

2.

3.

4. 1 - 2 + 4 - 8 +Á

+ s -1dn - 1 2n - 1+

Á

1 -

12

+

14

-

18

+Á

+ s -1dn - 1 1

2n - 1 +Á

9100

+

91002 +

91003 +

Á+

9100n +

Á

2 +

23

+

29

+

227

+Á

+

23n - 1 +

Á

5.

6.

Series with Geometric TermsIn Exercises 7–14, write out the first few terms of each series to showhow the series starts. Then find the sum of the series.

7. 8. aq

n = 2 14na

q

n = 0 s -1dn

4n

51 # 2

+

52 # 3

+

53 # 4

+Á

+

5nsn + 1d

+Á

12 # 3

+

13 # 4

+

14 # 5

+Á

+

1sn + 1dsn + 2d

+Á



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9. 10.

11. 12.

13. 14.

Telescoping SeriesUse partial fractions to find the sum of each series in Exercises 15–22.

15. 16.

17. 18.

19. 20.

21.

22.

Convergence or DivergenceWhich series in Exercises 23–40 converge, and which diverge? Givereasons for your answers. If a series converges, find its sum.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

Geometric SeriesIn each of the geometric series in Exercises 41–44, write out the firstfew terms of the series to find a and r, and find the sum of the series.

aq

n = 0 enp

pneaq

n = 0 a ep b

n

aq

n = 1 ln a n

2n + 1ba

q

n = 1 ln a n

n + 1b

aq

n = 1 nn

n!aq

n = 0

n!1000n

aq

n = 1 a1 -

1n b

n

aq

n = 0 2n

- 13n

aq

n = 0 1xn , ƒ x ƒ 7 1a

q

n = 1

210n

aq

n = 1 ln

1na

q

n = 0 e-2n

aq

n = 0 cos np

5naq

n = 0 cos np

aq

n = 1s -1dn + 1na

q

n = 1s -1dn + 1

32n

aq

n = 0A22 Bna

q

n = 0 a 122

bn

aq

n = 1stan-1 snd - tan-1 sn + 1dd

aq

n = 1 a 1

ln sn + 2d-

1ln sn + 1d

baq

n = 1 a 1

21>n -

1

21>sn + 1dba

q

n = 1 a 12n

-

12n + 1b

aq

n = 1

2n + 1n2sn + 1d2a

q

n = 1

40n

s2n - 1d2s2n + 1d2

aq

n = 1

6s2n - 1ds2n + 1da

q

n = 1

4s4n - 3ds4n + 1d

aq

n = 0 a2n + 1

5n baq

n = 0 a 1

2n +

s -1dn

5n baq

n = 0 a 5

2n -

13n ba

q

n = 0 a 5

2n +

13n b

aq

n = 0s -1dn

54na

q

n = 1 74n

Then express the inequality in terms of x and find the valuesof x for which the inequality holds and the series converges.

41. 42.

43. 44.

In Exercises 45–50, find the values of x for which the given geometricseries converges. Also, find the sum of the series (as a function of x)for those values of x.

45. 46.

47. 48.

49. 50.

Repeating DecimalsExpress each of the numbers in Exercises 51–58 as the ratio of twointegers.

51.

52.

53.

54. d is a digit

55.

56.

57.

58.

Theory and Examples59. The series in Exercise 5 can also be written as

Write it as a sum beginning with (a) (b)(c)

60. The series in Exercise 6 can also be written as

Write it as a sum beginning with (a) (b)(c)

61. Make up an infinite series of nonzero terms whose sum is

a. 1 b. c. 0.

62. (Continuation of Exercise 61.) Can you make an infinite series ofnonzero terms that converges to any number you want? Explain.

63. Show by example that may diverge even though and converge and no equals 0.bngbn

gangsan>bnd

-3

n = 20.n = 3,n = -1,

aq

n = 1

5nsn + 1d

and aq

n = 0

5sn + 1dsn + 2d

.

n = 5.n = 0,n = -2,

aq

n = 1

1sn + 1dsn + 2d

and aq

n = -1

1sn + 3dsn + 4d

.

3.142857 = 3.142857 142857 Á

1.24123 = 1.24 123 123 123 Á

1.414 = 1.414 414 414 Á

0.06 = 0.06666 Á

0.d = 0.dddd Á , where

0.7 = 0.7777 Á

0.234 = 0.234 234 234 Á

0.23 = 0.23 23 23 Á

aq

n = 0sln xdna

q

n = 0 sinn x

aq

n = 0 a- 1

2bn

sx - 3dnaq

n = 0s -1dnsx + 1dn

aq

n = 0s -1dnx-2na

q

n = 02nxn

aq

n = 0 s -1dn

2 a 1

3 + sin xbn

aq

n = 03 ax - 1

2bn

aq

n = 0s -1dnx2na

q

n = 0s -1dnxn

ƒ r ƒ 6 1




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64. Find convergent geometric series and thatillustrate the fact that may converge without being equalto AB.

65. Show by example that may converge to something otherthan A B even when and no equals 0.

66. If converges and for all n, can anything be said aboutGive reasons for your answer.

67. What happens if you add a finite number of terms to a divergentseries or delete a finite number of terms from a divergent series?Give reasons for your answer.

68. If converges and diverges, can anything be said abouttheir term-by-term sum Give reasons for your answer.

69. Make up a geometric series that converges to the number5 if

a. b.

70. Find the value of b for which

71. For what values of r does the infinite series

converge? Find the sum of the series when it converges.

72. Show that the error obtained by replacing a convergentgeometric series with one of its partial sums is

73. A ball is dropped from a height of 4 m. Each time it strikes the pave-ment after falling from a height of h meters it rebounds to a height of0.75h meters. Find the total distance the ball travels up and down.

74. (Continuation of Exercise 73.) Find the total number of secondsthe ball in Exercise 73 is traveling. (Hint: The formula gives )

75. The accompanying figure shows the first five of a sequence ofsquares. The outermost square has an area of Each of theother squares is obtained by joining the midpoints of the sides ofthe squares before it. Find the sum of the areas of all the squares.

76. The accompanying figure shows the first three rows and part ofthe fourth row of a sequence of rows of semicircles. There are semicircles in the nth row, each of radius Find the sum ofthe areas of all the semicircles.

1>2n .2n

4 m2 .

t = 2s>4.9 .s = 4.9t2

arn>s1 - rd .sn

sL - snd

1 + 2r + r2+ 2r3

+ r4+ 2r5

+ r6+

Á

1 + eb+ e2b

+ e3b+

Á= 9.

a = 13>2.a = 2

garn - 1

gsan + bnd?gbngan

gs1>and?an 7 0gan

bnA = gan, B = gbn Z 0,> gsan>bnd

gan bn

B = gbnA = gan

77. Helga von Koch’s snowflake curve Helga von Koch’s snow-flake is a curve of infinite length that encloses a region of finitearea. To see why this is so, suppose the curve is generated bystarting with an equilateral triangle whose sides have length 1.

a. Find the length of the nth curve and show that

b. Find the area of the region enclosed by and calculate

78. The accompanying figure provides an informal proof thatis less than 2. Explain what is going on. (Source:

“Convergence with Pictures” by P. J. Rippon, American Mathe-matical Monthly, Vol. 93, No. 6, 1986, pp. 476–478.)

11

1 …

…

12

132

122

142

152

14

162

172

gq

n=1 s1>n2d

Curve 1

Curve 4Curve 3

Curve 2

limn:q An .CnAn

limn:q Ln = q .CnLn

1/2

1/4

1/8




The Integral Test

Given a series we have two questions:

1. Does the series converge?

2. If it converges, what is its sum?

Much of the rest of this chapter is devoted to the first question, and in this section we answer thatquestion by making a connection to the convergence of the improper integral How-ever, as a practical matter the second question is also important, and we will return to it later.

In this section and the next two, we study series that do not have negative terms. Thereason for this restriction is that the partial sums of these series form nondecreasingsequences, and nondecreasing sequences that are bounded from above always converge(Theorem 6, Section 11.1). To show that a series of nonnegative terms converges, we needonly show that its partial sums are bounded from above.

It may at first seem to be a drawback that this approach establishes the fact of conver-gence without producing the sum of the series in question. Surely it would be better tocompute sums of series directly from formulas for their partial sums. But in most casessuch formulas are not available, and in their absence we have to turn instead to the two-step procedure of first establishing convergence and then approximating the sum.

Nondecreasing Partial Sums

Suppose that is an infinite series with for all n. Then each partial sum isgreater than or equal to its predecessor because

Since the partial sums form a nondecreasing sequence, the Nondecreasing Sequence The-orem (Theorem 6, Section 11.1) tells us that the series will converge if and only if the par-tial sums are bounded from above.

s1 … s2 … s3 …Á

… sn … sn + 1 …Á .

sn + 1 = sn + an :an Ú 0gq

n=1 an

1q

1 ƒsxd dx .

gan ,

11.3

Corollary of Theorem 6A series of nonnegative terms converges if and only if its partial sumsare bounded from above.

gq

n=1 an

EXAMPLE 1 The Harmonic Series

The series

is called the harmonic series. The harmonic series is divergent, but this doesn’t followfrom the nth-Term Test. The nth term 1 n does go to zero, but the series still diverges. Thereason it diverges is because there is no upper bound for its partial sums. To see why,group the terms of the series in the following way:

7 816 =

127 48 =

127 24 =

12

('''''')''''''*(''''')'''''*('')''*

1 +12

+ a13

+14b + a15 +

16

+17 +

18b + a1

9+

110

+Á

+116b +

Á.

>aq

n = 1 1n = 1 +

12

+13

+Á

+1n +

Á


Nicole Oresme(1320–1382)



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11.3 The Integral Test 773

The sum of the first two terms is 1.5. The sum of the next two terms is whichis greater than The sum of the next four terms is

which is greater than The sum of the nexteight terms is which isgreater than The sum of the next 16 terms is greater than andso on. In general, the sum of terms ending with is greater than The sequence of partial sums is not bounded from above: If the partial sum isgreater than k 2. The harmonic series diverges.

The Integral Test

We introduce the Integral Test with a series that is related to the harmonic series, butwhose nth term is instead of 1 n.

EXAMPLE 2 Does the following series converge?

Solution We determine the convergence of by comparing it withTo carry out the comparison, we think of the terms of the series as values of

the function and interpret these values as the areas of rectangles under thecurve

As Figure 11.7 shows,

Thus the partial sums of are bounded from above (by 2) and the seriesconverges. The sum of the series is known to be (See Exercise 16 inSection 11.11.)

p2>6 L 1.64493.gq

n=11>n2

6 1 + 1 = 2.

6 1 + Lq

1 1x2 dx

6 ƒs1d + Ln

1 1x2 dx

= ƒs1d + ƒs2d + ƒs3d +Á

+ ƒsnd

sn =112 +

122 +

132 +

Á+

1n2

y = 1>x2 .ƒsxd = 1>x2

1q

1 s1>x2d dx .gq

n=1s1>n2d

aq

n = 1 1n2 = 1 +

14

+19

+116

+Á

+1n2 +

Á

>1>n2

> snn = 2k ,2n>2n + 1

= 1>2.1>2n + 12n16>32 = 1>2,8>16 = 1>2.

1>15 + 1>16,1>9 + 1>10 + 1>11 + 1>12 + 1>13 + 1>14 +

1>8 + 1>8 + 1>8 + 1>8 = 1>2.1>7 + 1>8,1>5 + 1>6 +1>4 + 1>4 = 1>2.

1>3 + 1>4,

0 1

Graph of f(x) �

(1, f(1))

(2, f(2))

(3, f(3))(n, f(n))

2 3 4 … n � 1 n …

1x2

1n2

122

112

132

142

x

y

FIGURE 11.7 The sum of the areas of therectangles under the graph of is less than the area under the graph(Example 2).

f (x) = 1>x2

As in Section 8.8, Example 3,

1q

1 s1>x2d dx = 1 .

THEOREM 9 The Integral TestLet be a sequence of positive terms. Suppose that where ƒ is acontinuous, positive, decreasing function of x for all (N a positive inte-ger). Then the series and the integral both converge or bothdiverge.

1q

N ƒsxd dxgq

n=N an

x Ú Nan = ƒsnd ,5an6

Proof We establish the test for the case The proof for general N is similar.We start with the assumption that ƒ is a decreasing function with for every

n. This leads us to observe that the rectangles in Figure 11.8a, which have areasƒsnd = an

N = 1.

CautionThe series and integral need not have thesame value in the convergent case. As wenoted in Example 2,

while 1q

1 s1>x2d dx = 1.p2>6gq

n=1s1>n2d =



collectively enclose more area than that under the curve fromto That is,

In Figure 11.8b the rectangles have been faced to the left instead of to the right. If we mo-mentarily disregard the first rectangle, of area we see that

If we include we have

Combining these results gives

These inequalities hold for each n, and continue to hold as

If is finite, the right-hand inequality shows that is finite. If

is infinite, the left-hand inequality shows that is infinite. Hence the series

and the integral are both finite or both infinite.

EXAMPLE 3 The p-Series

Show that the p-series

( p a real constant) converges if and diverges if

Solution If then is a positive decreasing function of x. Since

the series converges by the Integral Test. We emphasize that the sum of the p-series is notThe series converges, but we don’t know the value it converges to.

If then and

The series diverges by the Integral Test.

Lq

1 1xp dx =

11 - p

limb: q

sb1 - p- 1d = q .

1 - p 7 0p 6 1,1>s p - 1d .

=1

1 - p s0 - 1d =

1p - 1

,

=1

1 - p limb: q

a 1b p - 1 - 1b

Lq

1 1xp dx = L

q

1 x-p dx = lim

b: q

c x-p + 1

-p + 1d

1

b

ƒsxd = 1>xpp 7 1,

p … 1.p 7 1,

aq

n = 1 1np =

11p +

12p +

13p +

Á+

1np +

Á

gan1q

1 ƒsxd dx

gan1q

1 ƒsxd dx

n : q .

Ln + 1

1 ƒsxd dx … a1 + a2 +

Á+ an … a1 + L

n

1 ƒsxd dx .

a1 + a2 +Á

+ an … a1 + Ln

1 ƒsxd dx .

a1 ,

a2 + a3 +Á

+ an … Ln

1 ƒsxd dx .

a1 ,

Ln + 1

1 ƒsxd dx … a1 + a2 +

Á+ an .

x = n + 1.x = 1y = ƒsxda1, a2, Á , an ,


0 1 2 n3 n � 1

a1a2

an

(a)

0 1 2 n3 n � 1

a1

a3an

(b)

a2

x

y

x

y

y � f (x)

y � f (x)

FIGURE 11.8 Subject to the conditions ofthe Integral Test, the series andthe integral both converge orboth diverge.

1q

1 ƒsxd dxgq

n=1 an

because p - 1 7 0.b p - 1 : q as b : q




If we have the (divergent) harmonic series

We have convergence for but divergence for every other value of p.

The p-series with is the harmonic series (Example 1). The p-Series Test showsthat the harmonic series is just barely divergent; if we increase p to 1.000000001, for in-stance, the series converges!

The slowness with which the partial sums of the harmonic series approaches infinityis impressive. For instance, it takes about 178,482,301 terms of the harmonic series tomove the partial sums beyond 20. It would take your calculator several weeks to compute asum with this many terms. (See also Exercise 33b.)

EXAMPLE 4 A Convergent Series

The series

converges by the Integral Test. The function is positive, continuous,and decreasing for and

Again we emphasize that is not the sum of the series. The series converges, but we donot know the value of its sum.

Convergence of the series in Example 4 can also be verified by comparison with theseries Comparison tests are studied in the next section.g1>n2 .

p>4 =

p2

-

p4

=

p4

.

= limb: q

[arctan b - arctan 1]

Lq

1

1x2

+ 1 dx = lim

b: q

Carctan x D1bx Ú 1,

ƒsxd = 1>sx2+ 1d

aq

n = 1

1n2

+ 1

p = 1

p 7 1

1 +12

+13

+Á

+1n +

Á .

p = 1,




EXERCISES 11.3

Determining Convergence or DivergenceWhich of the series in Exercises 1–30 converge, and which diverge?Give reasons for your answers. (When you check an answer, remem-ber that there may be more than one way to determine the series’ con-vergence or divergence.)

1. 2. 3.

4. 5. 6.

7. 8. 9. aq

n = 2 ln nna

q

n = 1 -8na

q

n = 1-

18n

aq

n = 1

-2

n2naq

n = 1

32naq

n = 1

5n + 1

aq

n = 1

nn + 1a

q

n = 1 e-na

q

n = 1

110n

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20.

21. 22. aq

n = 1

1ns1 + ln2 nda

q

n = 3

s1>nd

sln nd2ln2 n - 1

aq

n = 1

1sln 3dna

q

n = 1

1sln 2dn

aq

n = 1 a1 +

1n b

n

aq

n = 2 2nln na

q

n = 1

12n A2n + 1 B

aq

n = 1

2n

n + 1aq

n = 1

12n - 1a

q

n = 0

-2n + 1

aq

n = 1

5n

4n+ 3a

q

n = 1 2n

3naq

n = 2 ln n2n



tcu1103a.html

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tcu1103a.html

23. 24.

25. 26.

27. 28.

29. 30.

Theory and ExamplesFor what values of a, if any, do the series in Exercises 31 and 32converge?

31. 32.

33. a. Draw illustrations like those in Figures 11.7 and 11.8 to show thatthe partial sums of the harmonic series satisfy the inequalities

b. There is absolutely no empirical evidence for the divergenceof the harmonic series even though we know it diverges. Thepartial sums just grow too slowly. To see what we mean,suppose you had started with the day the universe wasformed, 13 billion years ago, and added a new term everysecond. About how large would the partial sum be today,assuming a 365-day year?

34. Are there any values of x for which converges?Give reasons for your answer.

35. Is it true that if is a divergent series of positive numbersthen there is also a divergent series of positive numberswith for every n? Is there a “smallest” divergent series ofpositive numbers? Give reasons for your answers.

36. (Continuation of Exercise 35.) Is there a “largest” convergent se-ries of positive numbers? Explain.

37. The Cauchy condensation test The Cauchy condensation testsays: Let be a nonincreasing sequence ( for all n)of positive terms that converges to 0. Then converges if andonly if converges. For example, diverges because

diverges. Show why the test works.

38. Use the Cauchy condensation test from Exercise 37 to show that

a. diverges;

b. converges if and diverges if p … 1.p 7 1aq

n = 1 1np

aq

n = 2

1n ln n

g2n # s1>2nd = g1gs1>ndg2na2n

gan

an Ú an + 15an6

bn 6 an

gq

n=1 bn

gq

n=1 an

gq

n=1s1>snxdd

sn

s1 = 1

… 1 + Ln

1 1x dx = 1 + ln n .

ln sn + 1d = Ln + 1

1 1x dx … 1 +

12

+Á

+

1n

aq

n = 3 a 1

n - 1-

2an + 1

baq

n = 1 a a

n + 2-

1n + 4

b

aq

n = 1 sech2 na

q

n = 1 sech n

aq

n = 1

n

n2+ 1a

q

n = 1 8 tan-1 n

1 + n2

aq

n = 1

21 + ena

q

n = 1

en

1 + e2n

aq

n = 1 n tan

1na

q

n = 1 n sin

1n

39. Logarithmic p-series

a. Show that

converges if and only if

b. What implications does the fact in part (a) have for theconvergence of the series

Give reasons for your answer.

40. (Continuation of Exercise 39.) Use the result in Exercise 39 to de-termine which of the following series converge and which di-verge. Support your answer in each case.

a. b.

c. d.

41. Euler’s constant Graphs like those in Figure 11.8 suggest that asn increases there is little change in the difference between the sum

and the integral

To explore this idea, carry out the following steps.

a. By taking in the proof of Theorem 9, show that

or

Thus, the sequence

is bounded from below and from above.

b. Show that

and use this result to show that the sequence in part (a)is decreasing.

5an6

1n + 1

6 Ln + 1

n 1x dx = ln sn + 1d - ln n ,

an = 1 +

12

+Á

+

1n - ln n

0 6 ln sn + 1d - ln n … 1 +

12

+Á

+

1n - ln n … 1.

ln sn + 1d … 1 +

12

+Á

+

1n … 1 + ln n

ƒsxd = 1>x

ln n = Ln

1 1x dx .

1 +

12

+Á

+

1n

aq

n = 2

1nsln nd3a

q

n = 2

1n ln sn3d

aq

n = 2

1nsln nd1.01a

q

n = 2

1nsln nd

aq

n = 2

1nsln nd p ?

p 7 1.

Lq

2

dxxsln xd p s p a positive constantd


T



tcu1103a.html

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777

Since a decreasing sequence that is bounded from below con-verges (Exercise 107 in Section 11.1), the numbers defined inpart (a) converge:

The number whose value is is called Euler’s con-stant. In contrast to other special numbers like and e, no otherp

0.5772 Á ,g ,

1 +

12

+Á

+

1n - ln n : g .

an

expression with a simple law of formulation has ever been foundfor

42. Use the integral test to show that

converges.

aq

n = 0e-n2

g .


11.3 The Integral Test


11.4 Comparison Tests 777

Comparison Tests

We have seen how to determine the convergence of geometric series, p-series, and a fewothers. We can test the convergence of many more series by comparing their terms to thoseof a series whose convergence is known.

11.4

THEOREM 10 The Comparison TestLet be a series with no negative terms.

(a) converges if there is a convergent series with for allfor some integer N.

(b) diverges if there is a divergent series of nonnegative terms withfor all for some integer N.n 7 N ,an Ú dn

gdngan

n 7 N ,an … cngcngan

gan

Proof In Part (a), the partial sums of are bounded above by

They therefore form a nondecreasing sequence with a limit In Part (b), the partial sums of are not bounded from above. If they were, the par-

tial sums for would be bounded by

and would have to converge instead of diverge.

EXAMPLE 1 Applying the Comparison Test

(a) The series

diverges because its nth term

is greater than the nth term of the divergent harmonic series.

55n - 1

=1

n -15

71n

aq

n = 1

55n - 1

gdn

M*= d1 + d2 +

Á+ dN + a

q

n = N + 1an

gdn

gan

L … M .

M = a1 + a2 +Á

+ aN + aq

n = N + 1cn .

ganHISTORICAL BIOGRAPHY

Albert of Saxony(ca. 1316–1390)



bounce11.html?3_0_a


(b) The series

converges because its terms are all positive and less than or equal to the correspon-ding terms of

The geometric series on the left converges and we have

The fact that 3 is an upper bound for the partial sums of does notmean that the series converges to 3. As we will see in Section 11.9, the series con-verges to e.

(c) The series

converges. To see this, we ignore the first three terms and compare the remaining termswith those of the convergent geometric series The term ofthe truncated sequence is less than the corresponding term of the geometric se-ries. We see that term by term we have the comparison,

So the truncated series and the original series converge by an application of the Com-parison Test.

The Limit Comparison Test

We now introduce a comparison test that is particularly useful for series in which is arational function of n.

an

1 +1

2 + 21+

1

4 + 22+

1

8 + 23+

Á… 1 +

12

+14

+18

+Á

1>2n1>s2n

+ 2ndgq

n=0 s1>2nd .

5 +23

+17 + 1 +

1

2 + 21+

1

4 + 22+

1

8 + 23+

Á+

1

2n+ 2n

+Á

gq

n=0 s1>n!d

1 + aq

n = 0 12n = 1 +

11 - s1>2d

= 3.

1 + aq

n = 0 12n = 1 + 1 +

12

+122 +

Á .

aq

n = 0 1n!

= 1 +11!

+12!

+13!

+Á

THEOREM 11 Limit Comparison TestSuppose that and for all (N an integer).

1. If then and both converge or both diverge.

2. If and converges, then converges.

3. If and diverges, then diverges.gangbnlimn: q

an

bn= q

gangbnlimn: q

an

bn= 0

gbnganlimn: q

an

bn= c 7 0,

n Ú Nbn 7 0an 7 0



11.4 Comparison Tests 779

Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b).Since there exists an integer N such that for all n

Thus, for

If converges, then converges and converges by the Direct Compari-son Test. If diverges, then diverges and diverges by the Direct Com-parison Test.

EXAMPLE 2 Using the Limit Comparison Test

Which of the following series converge, and which diverge?

(a)

(b)

(c)

Solution

(a) Let For large n, we expect to behave likesince the leading terms dominate for large n, so we let Since

and

diverges by Part 1 of the Limit Comparison Test. We could just as well havetaken but 1 n is simpler.>bn = 2>n ,gan

limn: q

an

bn= lim

n: q

2n2

+ nn2

+ 2n + 1= 2,

aq

n = 1bn = a

q

n = 1 1n diverges

bn = 1>n .2n>n2= 2>n anan = s2n + 1d>sn2

+ 2n + 1d .

1 + 2 ln 29

+

1 + 3 ln 314

+1 + 4 ln 4

21+

Á= a

q

n = 2 1 + n ln n

n2+ 5

11

+13

+17 +

115

+Á

= aq

n = 1

12n

- 1

34

+

59

+

716

+

925

+Á

= aq

n = 1

2n + 1sn + 1d2 = a

q

n = 1

2n + 1n2

+ 2n + 1

gangsc>2dbngbn

gangs3c>2dbngbn

ac2bbn 6 an 6 a3c

2bbn .

c2

6

an

bn6

3c2

,

-c2

6

an

bn- c 6

c2

,

n 7 N ,

n 7 N Q ` an

bn- c ` 6

c2

.

c>2 7 0,Limit definition with

andreplaced by an>bnan

P = c>2, L = c ,



(b) Let For large n, we expect to behave like so we letSince

and

converges by Part 1 of the Limit Comparison Test.

(c) Let For large n, we expect to behave likewhich is greater than 1 n for so we take

Since

and

diverges by Part 3 of the Limit Comparison Test.

EXAMPLE 3 Does converge?

Solution Because ln n grows more slowly than for any positive constant c(Section 11.1, Exercise 91), we would expect to have

for n sufficiently large. Indeed, taking and we have

Since (a p-series with ) converges, converges by Part 2 ofthe Limit Comparison Test.

ganp 7 1gbn = gs1>n5>4d

= limn: q

4

n1>4 = 0.

= limn: q

1>n

s1>4dn-3>4

limn: q

an

bn= lim

n: q

ln n

n1>4

bn = 1>n5>4 ,an = sln nd>n3>2

ln n

n3>2 6

n1>4n3>2 =

1n5>4

nc

aq

n = 1 ln n

n3>2

gan

= q ,

limn: q

an

bn= lim

n: q

n + n2 ln n

n2+ 5

aq

n = 2bn = a

q

n = 2 1n diverges

bn = 1>n .n Ú 3,>sn ln nd>n2= sln nd>n ,

anan = s1 + n ln nd>sn2+ 5d .

gan

= 1,

= limn: q

1

1 - s1>2nd

limn: q

an

bn= lim

n: q

2n

2n- 1

aq

n = 1bn = a

q

n = 1 12n converges

bn = 1>2n .1>2n ,anan = 1>s2n

- 1d .


l’Hôpital’s Rule



781

EXERCISES 11.4

Determining Convergence or DivergenceWhich of the series in Exercises 1–36 converge, and which diverge?Give reasons for your answers.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. 26.

27. 28.

29. 30. 31.

32. 33. 34.

35. 36.

Theory and Examples37. Prove (a) Part 2 and (b) Part 3 of the Limit Comparison Test.

aq

n = 1

11 + 22

+ 32+

Á+ n2a

q

n = 1

11 + 2 + 3 +

Á+ n

aq

n = 1 2n n

n2aq

n = 1

1

n2n naq

n = 1 tanh n

n2

aq

n = 1 coth n

n2aq

n = 1 sec-1 n

n1.3aq

n = 1 tan-1 n

n1.1

aq

n = 3

5n3- 3n

n2sn - 2dsn2+ 5da

q

n = 1

10n + 1nsn + 1dsn + 2d

aq

n = 1 tan

1na

q

n = 1 sin

1n

aq

n = 1 3n - 1

+ 13na

q

n = 1

13n - 1

+ 1aq

n = 1 n + 2n

n22n

aq

n = 1 1 - n

n2naq

n = 1 2n

n2+ 1a

q

n = 2

1

n2n2- 1

aq

n = 1

1s1 + ln2 nda

q

n = 2 ln sn + 1d

n + 1aq

n = 1

1s1 + ln nd2

aq

n = 1

11 + ln na

q

n = 1 sln nd2

n3>2aq

n = 2

12n ln n

aq

n = 1 sln nd3

n3aq

n = 1 sln nd2

n3aq

n = 2

1sln nd2

aq

n = 3

1ln sln nda

q

n = 1

12n3+ 2

aq

n = 1 a n

3n + 1bn

aq

n = 1 n + 1

n22naq

n = 1

2n3n - 1a

q

n = 1 1 + cos n

n2

aq

n = 1 sin2 n

2naq

n = 1

3

n + 2naq

n = 1

1

22n + 23 n

38. If is a convergent series of nonnegative numbers, cananything be said about Explain.

39. Suppose that and for (N an integer). Ifand converges, can anything be said

about Give reasons for your answer.

40. Prove that if is a convergent series of nonnegative terms,then converges.

COMPUTER EXPLORATION

41. It is not yet known whether the series

converges or diverges. Use a CAS to explore the behavior of theseries by performing the following steps.

a. Define the sequence of partial sums

What happens when you try to find the limit of as Does your CAS find a closed form answer for this limit?

b. Plot the first 100 points for the sequence of partialsums. Do they appear to converge? What would you estimatethe limit to be?

c. Next plot the first 200 points Discuss the behavior inyour own words.

d. Plot the first 400 points What happens whenCalculate the number 355 113. Explain from your

calculation what happened at For what values of kwould you guess this behavior might occur again?

You will find an interesting discussion of this series in Chapter 72of Mazes for the Mind by Clifford A. Pickover, St. Martin’s Press,Inc., New York, 1992.

k = 355.>k = 355?

sk, skd .

sk, skd .

sk, skd

k : q ?sk

sk = ak

n = 1

1n3 sin2 n

.

aq

n = 1

1n3 sin2 n

gan2gan

g bn ?ganlim n:q san>bnd = q

n Ú Nbn 7 0an 7 0

gq

n=1san>nd?gq

n=1 an


11.4 Comparison Tests


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11.5 The Ratio and Root Tests 781

The Ratio and Root Tests

The Ratio Test measures the rate of growth (or decline) of a series by examining the ratioFor a geometric series this rate is a constant and the

series converges if and only if its ratio is less than 1 in absolute value. The Ratio Test is apowerful rule extending that result. We prove it on the next page using the Comparison Test.

ssarn + 1d>sarnd = rd ,garn ,an + 1>an .

11.5



Proof

(a) Let r be a number between and 1. Then the number is positive.Since

must lie within of when n is large enough, say for all In particular

That is,

These inequalities show that the terms of our series, after the Nth term, approach zeromore rapidly than the terms in a geometric series with ratio More precisely,consider the series where for and

Now for all n, and

The geometric series converges because so con-verges. Since also converges.

(b) From some index M on,

The terms of the series do not approach zero as n becomes infinite, and the seriesdiverges by the nth-Term Test.

an + 1an

7 1 and aM 6 aM + 1 6 aM + 2 6Á .

1<R ◊ ˆ .

an … cn, gan

gcnƒ r ƒ 6 1,1 + r + r2+

Á

= a1 + a2 +Á

+ aN - 1 + aN s1 + r + r2+

Ád .

aq

n = 1cn = a1 + a2 +

Á+ aN - 1 + aN + raN + r2aN +

Á

an … cnr2aN, Á , cN + m = rmaN, Á .cN + 1 = raN, cN + 2 =n = 1, 2, Á , Ncn = angcn ,

r 6 1.

aN + m 6 raN + m - 1 6 r maN .

o

aN + 3 6 raN + 2 6 r 3aN ,

aN + 2 6 raN + 1 6 r 2aN ,

aN + 1 6 raN ,

an + 1an

6 r + P = r, when n Ú N .

n Ú N .rPan + 1>an

an + 1an

: r ,

P = r - rrR<1.


THEOREM 12 The Ratio TestLet be a series with positive terms and suppose that

Then

(a) the series converges if ,

(b) the series diverges if or is infinite,

(c) the test is inconclusive if r = 1.

rr 7 1

r 6 1

limn: q

an + 1an

= r .

gan




(c) The two series

show that some other test for convergence must be used when

In both cases, yet the first series diverges, whereas the second converges.

The Ratio Test is often effective when the terms of a series contain factorials of ex-pressions involving n or expressions raised to a power involving n.

EXAMPLE 1 Applying the Ratio Test

Investigate the convergence of the following series.

(a) (b) (c)

Solution

(a) For the series

The series converges because is less than 1. This does not mean that 2 3 isthe sum of the series. In fact,

(b) If then and

The series diverges because is greater than 1.

(c) If then

=

4sn + 1dsn + 1ds2n + 2ds2n + 1d

=

2sn + 1d2n + 1

: 1.

an + 1an

=

4n + 1sn + 1d!sn + 1d!s2n + 2ds2n + 1ds2nd!

#s2nd!4nn!n!

an = 4nn!n!>s2nd! ,

r = 4

=

s2n + 2ds2n + 1dsn + 1dsn + 1d

=

4n + 2n + 1

: 4.

an + 1an

=

n!n!s2n + 2ds2n + 1ds2nd!sn + 1d!sn + 1d!s2nd!

an + 1 =

s2n + 2d!sn + 1d!sn + 1d!

an =

s2nd!n!n!

,

aq

n = 0 2n

+ 53n = a

q

n = 0 a2

3bn

+ aq

n = 0 53n =

11 - s2>3d

+

51 - s1>3d

=212

.

>r = 2>3

an + 1an

=

s2n + 1+ 5d>3n + 1

s2n+ 5d>3n =

13

# 2n + 1

+ 52n

+ 5=

13

# a2 + 5 # 2-n

1 + 5 # 2-n b : 13

# 21

=23

.

gq

n=0 s2n+ 5d>3n ,

aq

n = 1 4nn!n!s2nd!a

q

n = 1 s2nd!n!n!a

q

n = 0 2n

+ 53n

r = 1,

For aq

n = 1 1n2: an + 1

an=

1>sn + 1d2

1>n2 = a nn + 1

b2

: 12= 1.

For aq

n = 1 1n:

an + 1an

=

1>sn + 1d1>n =

nn + 1

: 1.

r = 1.

aq

n = 1 1n and a

q

n = 1 1n2

R = 1.



Because the limit is we cannot decide from the Ratio Test whether the seriesconverges. When we notice that we conclude that

is always greater than because is always greater than 1.Therefore, all terms are greater than or equal to and the nth term does not ap-proach zero as The series diverges.

The Root Test

The convergence tests we have so far for work best when the formula for is rela-tively simple. But consider the following.

EXAMPLE 2 Let Does converge?

Solution We write out several terms of the series:

Clearly, this is not a geometric series. The nth term approaches zero as so we donot know if the series diverges. The Integral Test does not look promising. The Ratio Testproduces

As the ratio is alternately small and large and has no limit.A test that will answer the question (the series converges) is the Root Test.

n : q ,

an + 1an

= d 12n

, n odd

n + 12

, n even.

n : q ,

=12

+14

+

38

+116

+

532

+1

64+

7128

+Á .

aq

n = 1an =

121 +

122 +

323 +

124 +

525 +

126 +

727 +

Á

ganan = en>2n, n odd

1>2n, n even.

angan

n : q .a1 = 2,

s2n + 2d>s2n + 1danan + 1

an + 1>an = s2n + 2d>s2n + 1d ,r = 1,


THEOREM 13 The Root TestLet be a series with for and suppose that

Then

(a) the series converges if

(b) the series diverges if or is infinite,

(c) the test is inconclusive if r = 1.

rr 7 1

r 6 1,

limn: q

2n an = r .

n Ú N ,an Ú 0gan

Proof

(a) Choose an so small that Since the terms eventually get closer than to In other words, there exists an index suchthat 2n an 6 r + P when n Ú M .

M Ú Nr .P

2n an2n an : r ,r + P 6 1.P 7 0R<1.




Then it is also true that

Now, a geometric series with ratio converges. Bycomparison, converges, from which it follows that

converges.

(b) For all indices beyond some integer M, we have so thatfor The terms of the series do not converge to zero. The series di-

verges by the nth-Term Test.

(c) The series and show that the test is not conclusivewhen The first series diverges and the second converges, but in both cases

EXAMPLE 3 Applying the Root Test

Which of the following series converges, and which diverges?

(a) (b) (c)

Solution

(a) converges because

(b) diverges because

(c) converges because

EXAMPLE 2 Revisited

Let Does converge?

Solution We apply the Root Test, finding that

Therefore,

Since (Section 11.1, Theorem 5), we have by the SandwichTheorem. The limit is less than 1, so the series converges by the Root Test.

limn:q2n an = 1>22n n : 1

12

… 2n an …

2n n2

.

2n an = e2n n>2, n odd 1>2, n even.

ganan = en>2n, n odd

1>2n, n even.

Bn a 11 + n

bn

=1

1 + n : 0 6 1.a

q

n = 1 a 1

1 + nbn

An 2n

n2 =2

A2n n B2 : 21

7 1.aq

n = 1 2n

n2

Bn n2

2n =

2n n22n 2n=

A2n n B22

: 12

6 1.aq

n = 1 n2

2n

aq

n = 1 a 1

1 + nbn

aq

n = 1 2n

n2aq

n = 1 n2

2n

2n an : 1.r = 1.

gq

n=1 s1>n2dgq

n=1 s1>ndR = 1.

n 7 M .an 7 12n an 7 1,1<R ◊ ˆ .

aq

n = 1an = a1 +

Á+ aM - 1 + a

q

n = Man

gq

n=M an

sr + Pd 6 1,gq

n=M sr + Pdn ,

an 6 sr + Pdn for n Ú M .




EXERCISES 11.5

Determining Convergence or DivergenceWhich of the series in Exercises 1–26 converge, and which diverge?Give reasons for your answers. (When checking your answers, remem-ber there may be more than one way to determine a series’ conver-gence or divergence.)

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

Which of the series defined by the formulas in Exercises27–38 converge, and which diverge? Give reasons for your answers.

27.

28.

29.

30.

31. a1 = 2, an + 1 =

2n an

a1 = 3, an + 1 =

nn + 1

an

a1 =

13

, an + 1 =

3n - 12n + 5

an

a1 = 1, an + 1 =

1 + tan-1 nn an

a1 = 2, an + 1 =

1 + sin nn an

gq

n=1 an

aq

n = 1

3n

n32naq

n = 1

n! ln nnsn + 2d!

aq

n = 2

n

sln ndsn>2daq

n = 2

nsln ndn

aq

n = 1 n!nna

q

n = 1

n!s2n + 1d!

aq

n = 1 n2nsn + 1d!

3nn!aq

n = 1 sn + 3d!

3!n!3n

aq

n = 1e-nsn3da

q

n = 1 sn + 1dsn + 2d

n!

aq

n = 1 n ln n

2naq

n = 1 ln nn

aq

n = 1 a1n -

1n2 b

n

aq

n = 1 a1n -

1n2 b

aq

n = 1 sln ndn

nnaq

n = 1 ln n

n3

aq

n = 1 a1 -

13nbn

aq

n = 1 a1 -

3n b

n

aq

n = 1 s -2dn

3naq

n = 1 2 + s -1dn

1.25n

aq

n = 1 an - 2

n bn

aq

n = 1 n10

10n

aq

n = 1

n!10na

q

n = 1n!e-n

aq

n = 1n2e-na

q

n = 1 n22

2n

32.

33.

34.

35.

36.

37.

38.

Which of the series in Exercises 39–44 converge, and which diverge?Give reasons for your answers.

39. 40.

41. 42.

43.

44.

Theory and Examples45. Neither the Ratio nor the Root Test helps with p-series. Try them

on

and show that both tests fail to provide information about conver-gence.

46. Show that neither the Ratio Test nor the Root Test provides infor-mation about the convergence of

47. Let

Does converge? Give reasons for your answer.gan

an = en>2n, if n is a prime number

1>2n, otherwise.

aq

n = 2

1sln nd p s p constantd .

aq

n = 1 1np

aq

n = 1

1 # 3 # Á # s2n - 1d[2 # 4 # Á # s2nd]s3n

+ 1d

aq

n = 1 1 # 3 # Á # s2n - 1d

4n2nn!

aq

n = 1

nn

s2nd2aq

n = 1

nn

2sn2d

aq

n = 1 sn!dn

nsn2daq

n = 1 sn!dn

snnd2

an =

s3nd!n!sn + 1d!sn + 2d!

an =

2nn!n!s2nd!

a1 =

12

, an + 1 = sandn + 1

a1 =

13

, an + 1 = 2n an

a1 =

12

, an + 1 =

n + ln nn + 10

an

a1 = 1, an + 1 =

1 + ln nn an

a1 = 5, an + 1 =

2n n2

an



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11.6 Alternating Series, Absolute and Conditional Convergence 787

Alternating Series, Absolute and Conditional Convergence

A series in which the terms are alternately positive and negative is an alternating series.Here are three examples:

(1)

(2)

(3)

Series (1), called the alternating harmonic series, converges, as we will see in a moment.Series (2) a geometric series with ratio converges to Series (3) diverges because the nth term does not approach zero.

We prove the convergence of the alternating harmonic series by applying the AlternatingSeries Test.

-4>3.-2>[1 + s1>2d] =r = -1>2,

1 - 2 + 3 - 4 + 5 - 6 +Á

+ s -1dn + 1n +Á

-2 + 1 -12

+14

-18

+Á

+

s -1dn42n +

Á

1 -12

+13

-14

+15 -

Á+

s -1dn + 1

n +Á

11.6

THEOREM 14 The Alternating Series Test (Leibniz’s Theorem)The series

converges if all three of the following conditions are satisfied:

1. The are all positive.

2. for all for some integer N.

3. un : 0.

n Ú N ,un Ú un + 1

un’s

aq

n = 1s -1dn + 1un = u1 - u2 + u3 - u4 +

Á

Proof If n is an even integer, say then the sum of the first n terms is

The first equality shows that is the sum of m nonnegative terms, since each termin parentheses is positive or zero. Hence and the sequence is non-decreasing. The second equality shows that Since is nondecreasing andbounded from above, it has a limit, say

(4)

If n is an odd integer, say then the sum of the first n terms isSince

and, as

(5)

Combining the results of Equations (4) and (5) gives (Section 11.1, Exer-

cise 119).

limn: q

sn = L

s2m + 1 = s2m + u2m + 1 : L + 0 = L .

m : q ,

limm: q

u2m + 1 = 0

un : 0,s2m + 1 = s2m + u2m + 1 .n = 2m + 1,

limm: q

s2m = L .

5s2m6s2m … u1 .5s2m6s2m + 2 Ú s2m ,

s2m

= u1 - su2 - u3d - su4 - u5d -Á

- su2m - 2 - u2m - 1d - u2m .

s2m = su1 - u2d + su3 - u4d +Á

+ su2m - 1 - u2md

n = 2m ,




EXAMPLE 1 The alternating harmonic series

satisfies the three requirements of Theorem 14 with it therefore converges.

A graphical interpretation of the partial sums (Figure 11.9) shows how an alternatingseries converges to its limit L when the three conditions of Theorem 14 are satisfied with

(Exercise 63 asks you to picture the case ) Starting from the origin of thex-axis, we lay off the positive distance To find the point corresponding to

we back up a distance equal to Since we do not back up anyfarther than the origin. We continue in this seesaw fashion, backing up or going forward asthe signs in the series demand. But for each forward or backward step is shorterthan (or at most the same size as) the preceding step, because And since thenth term approaches zero as n increases, the size of step we take forward or backward getssmaller and smaller. We oscillate across the limit L, and the amplitude of oscillation ap-proaches zero. The limit L lies between any two successive sums and and hence dif-fers from by an amount less than

Because

we can make useful estimates of the sums of convergent alternating series.

ƒ L - sn ƒ 6 un + 1 for n Ú N ,

un + 1 .sn

sn + 1sn

un + 1 … un .n Ú N ,

u2 … u1 ,u2 .s2 = u1 - u2 ,s1 = u1 .

N 7 1.N = 1.

N = 1;

aq

n = 1s -1dn + 1

1n = 1 -

12

+13

-14

+Á

L0

�u1

�u2

�u3

�u4

s2 s4 s3 s1

x

FIGURE 11.9 The partial sums of analternating series that satisfies thehypotheses of Theorem 14 for straddle the limit from the beginning.

N = 1

THEOREM 15 The Alternating Series Estimation TheoremIf the alternating series satisfies the three conditions ofTheorem 14, then for

approximates the sum L of the series with an error whose absolute value is lessthan the numerical value of the first unused term. Furthermore, the remain-der, has the same sign as the first unused term.L - sn ,

un + 1 ,

sn = u1 - u2 +Á

+ s -1dn + 1un

n Ú N ,gq

n=1 s -1dn + 1un

We leave the verification of the sign of the remainder for Exercise 53.

EXAMPLE 2 We try Theorem 15 on a series whose sum we know:

The theorem says that if we truncate the series after the eighth term, we throw away a totalthat is positive and less than 1 256. The sum of the first eight terms is 0.6640625. The sumof the series is

The difference, is positive and less thans1>256d = 0.00390625.

s2>3d - 0.6640625 = 0.0026041666 Á ,

11 - s -1>2d

=1

3>2 =23

.

>

aq

n = 0s -1dn

12n = 1 -

12

+14

-18

+116

-132

+164

-1

128 +

1256

-Á .

----

--




Absolute and Conditional Convergence

DEFINITION Absolutely ConvergentA series converges absolutely (is absolutely convergent) if the correspon-ding series of absolute values, converges.g ƒ an ƒ ,

gan

The geometric series

converges absolutely because the corresponding series of absolute values

converges. The alternating harmonic series does not converge absolutely. The correspondingseries of absolute values is the (divergent) harmonic series.

1 +12

+14

+18

+Á

1 -12

+14

-18

+Á

DEFINITION Conditionally ConvergentA series that converges but does not converge absolutely converges conditionally.

THEOREM 16 The Absolute Convergence Test

If converges, then converges.aq

n = 1ana

q

n = 1 ƒ an ƒ

The alternating harmonic series converges conditionally.Absolute convergence is important for two reasons. First, we have good tests for con-

vergence of series of positive terms. Second, if a series converges absolutely, then it con-verges. That is the thrust of the next theorem.

Proof For each n,

If converges, then converges and, by the Direct Comparison Test,the nonnegative series converges. The equality now lets us express as the difference of two convergent series:

Therefore, converges.gq

n=1 an

aq

n = 1 an = a

q

n = 1san + ƒ an ƒ - ƒ an ƒ d = a

q

n = 1san + ƒ an ƒ d - a

q

n = 1ƒ an ƒ .

gq

n=1 an

an = san + ƒ an ƒ d - ƒ an ƒgq

n=1 san + ƒ an ƒ dgq

n=1 2 ƒ an ƒgq

n=1 ƒ an ƒ

- ƒ an ƒ … an … ƒ an ƒ, so 0 … an + ƒ an ƒ … 2 ƒ an ƒ .



CAUTION We can rephrase Theorem 16 to say that every absolutely convergent seriesconverges. However, the converse statement is false: Many convergent series do not con-verge absolutely (such as the alternating harmonic series in Example 1).

EXAMPLE 3 Applying the Absolute Convergence Test

(a) For the corresponding series of absolute

values is the convergent series

The original series converges because it converges absolutely.

(b) For the corresponding series of absolute

values is

which converges by comparison with because for every n.The original series converges absolutely; therefore it converges.

EXAMPLE 4 Alternating p-Series

If p is a positive constant, the sequence is a decreasing sequence with limit zero.Therefore the alternating p-series

converges.If the series converges absolutely. If the series converges condi-

tionally.

Rearranging Series

Absolute convergence: 1 -1

23>2 +1

33>2 -1

43>2 +Á

Conditional convergence: 1 -122

+123

-124

+Á

0 6 p … 1,p 7 1,

aq

n = 1 s -1dn - 1

np = 1 -12p +

13p -

14p +

Á , p 7 0

51>np6

ƒ sin n ƒ … 1gq

n=1 s1>n2d

aq

n = 1` sin n

n2 ` =

ƒ sin 1 ƒ

1+

ƒ sin 2 ƒ

4+

Á ,

aq

n = 1 sin nn2 =

sin 11

+

sin 24

+

sin 39

+Á ,

aq

n = 1 1n2 = 1 +

14

+19

+116

+Á .

aq

n = 1s -1dn + 1

1n2 = 1 -

14

+19

-116

+Á ,


THEOREM 17 The Rearrangement Theorem for AbsolutelyConvergent Series

If converges absolutely, and is any arrangement of thesequence then converges absolutely and

aq

n = 1bn = a

q

n = 1an .

gbn5an6 ,b1, b2 , Á , bn , Ágq

n=1 an

(For an outline of the proof, see Exercise 60.)




EXAMPLE 5 Applying the Rearrangement Theorem

As we saw in Example 3, the series

converges absolutely. A possible rearrangement of the terms of the series might start witha positive term, then two negative terms, then three positive terms, then four negativeterms, and so on: After k terms of one sign, take terms of the opposite sign. The firstten terms of such a series look like this:

The Rearrangement Theorem says that both series converge to the same value. In this ex-ample, if we had the second series to begin with, we would probably be glad to exchange itfor the first, if we knew that we could. We can do even better: The sum of either series isalso equal to

(See Exercise 61.)

If we rearrange infinitely many terms of a conditionally convergent series, we can getresults that are far different from the sum of the original series. Here is an example.

EXAMPLE 6 Rearranging the Alternating Harmonic Series

The alternating harmonic series

can be rearranged to diverge or to reach any preassigned sum.

(a) Rearranging to diverge. The series of terms di-verges to and the series of terms diverges to No matter how farout in the sequence of odd-numbered terms we begin, we can always add enough pos-itive terms to get an arbitrarily large sum. Similarly, with the negative terms, no matterhow far out we start, we can add enough consecutive even-numbered terms to get anegative sum of arbitrarily large absolute value. If we wished to do so, we could startadding odd-numbered terms until we had a sum greater than say, and then followthat with enough consecutive negative terms to make the new total less than Wecould then add enough positive terms to make the total greater than and followwith consecutive unused negative terms to make a new total less than and so on.In this way, we could make the swings arbitrarily large in either direction.

(b) Rearranging to converge to 1. Another possibility is to focus on aparticular limit. Suppose we try to get sums that converge to 1. We start with the firstterm, 1 1, and then subtract 1 2. Next we add 1 3 and 1 5, which brings the totalback to 1 or above. Then we add consecutive negative terms until the total is less than1. We continue in this manner: When the sum is less than 1, add positive terms untilthe total is 1 or more; then subtract (add negative) terms until the total is again lessthan 1. This process can be continued indefinitely. Because both the odd-numbered

>>>>gq

n=1 s -1dn + 1>n-6,+5

-4.+3,

- q .gs -1>2nd+ q

g[1>s2n - 1d]gq

n=1 s -1dn + 1>n

11

-12

+13

-14

+15 -

16

+17 -

18

+19

-110

+111

-Á

aq

n = 1

1s2n - 1d2 - a

q

n = 1

1s2nd2 .

1 -14

-116

+19

+125

+149

-1

36-

164

-1

100-

1144

+Á .

k + 1

1 -14

+19

-116

+Á

+ s -1dn - 1 1n2 +

Á



terms and the even-numbered terms of the original series approach zero as the amount by which our partial sums exceed 1 or fall below it approaches zero. So thenew series converges to 1. The rearranged series starts like this:

The kind of behavior illustrated by the series in Example 6 is typical of what can hap-pen with any conditionally convergent series. Therefore we must always add the terms of aconditionally convergent series in the order given.

We have now developed several tests for convergence and divergence of series. Insummary:

123

+1

25-

114

+127

-116

+Á

+119

+121

-1

12+

16

+111

+113

-18

+1

15+

117

-110

11

-12

+13

+15 -

14

+17 +

19

-

n : q ,


1. The nth-Term Test: Unless the series diverges.

2. Geometric series: converges if otherwise it diverges.

3. p-series: converges if otherwise it diverges.

4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or RootTest. Try comparing to a known series with the Comparison Test.

5. Series with some negative terms: Does converge? If yes, so doessince absolute convergence implies convergence.

6. Alternating series: converges if the series satisfies the conditions ofthe Alternating Series Test.

gan

gan ,g ƒ an ƒ

p 7 1;g1>np

ƒ r ƒ 6 1;garn

an : 0,




EXERCISES 11.6

Determining Convergence or DivergenceWhich of the alternating series in Exercises 1–10 converge, and whichdiverge? Give reasons for your answers.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

Absolute ConvergenceWhich of the series in Exercises 11–44 converge absolutely, whichconverge, and which diverge? Give reasons for your answers.

aq

n = 1s -1dn + 1

32n + 12n + 1aq

n = 1s -1dn + 1

2n + 1n + 1

aq

n = 1s -1dn ln a1 +

1n ba

q

n = 2s -1dn + 1

ln n

ln n2

aq

n = 1s -1dn + 1

ln nna

q

n = 2s -1dn + 1

1ln n

aq

n = 1s -1dn + 1

10n

n10aq

n = 1s -1dn + 1 a n

10bn

aq

n = 1s -1dn + 1

1

n3>2aq

n = 1s -1dn + 1

1n2

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26. aq

n = 2s -1dn + 1

1n ln na

q

n = 1s -1dn

tan-1 n

n2+ 1

aq

n = 1s -1dn + 1 A2n 10 Ba

q

n = 1s -1dnn2s2>3dn

aq

n = 1 s -2dn + 1

n + 5naq

n = 1s -1dn + 1

1 + n

n2

aq

n = 2s -1dn

1ln sn3da

q

n = 1s -1dn + 1

3 + n5 + n

aq

n = 1s -1dn

sin n

n2aq

n = 1s -1dn

1n + 3

aq

n = 1s -1dn + 1

n!2na

q

n = 1s -1dn + 1

n

n3+ 1

aq

n = 1

s -1dn

1 + 2naq

n = 1s -1dn

12n

aq

n = 1s -1dn + 1

s0.1dn

naq

n = 1s -1dn + 1s0.1dn



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27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41.

42. 43.

44.

Error EstimationIn Exercises 45–48, estimate the magnitude of the error involved inusing the sum of the first four terms to approximate the sum of the en-tire series.

45.

46.

47.

48.

Approximate the sums in Exercises 49 and 50 with an error of magni-tude less than

49.

50.

Theory and Examples51. a. The series

does not meet one of the conditions of Theorem 14. Which one?

b. Find the sum of the series in part (a).

13

-

12

+

19

-

14

+

127

-

18

+Á

+

13n -

12n +

Á

aq

n = 0s -1dn

1n!

aq

n = 0s -1dn

1s2nd!

5 * 10-6 .

11 + t

= aq

n = 0s -1dnt n, 0 6 t 6 1

aq

n = 1s -1dn + 1

s0.01dn

n

aq

n = 1s -1dn + 1

110n

aq

n = 1s -1dn + 1

1n

aq

n = 1s -1dn csch n

aq

n = 1s -1dn sech na

q

n = 1

s -1dn2n + 2n + 1

aq

n = 1s -1dn A2n + 1n - 2n B

aq

n = 1s -1dn A2n2

+ n - n Baq

n = 1s -1dn A2n + 1 - 2n B

aq

n = 1s -1dn

sn!d2 3n

s2n + 1d!aq

n = 1s -1dn

s2nd!2nn!n

aq

n = 1 s -1dn + 1sn!d2

s2nd!aq

n = 1 s -1dnsn + 1dn

s2ndn

aq

n = 1 cos np

naq

n = 1 cos np

n2n

aq

n = 2s -1dn a ln n

ln n2 bn

aq

n = 1

s -1dn - 1

n2+ 2n + 1

aq

n = 1s -5d-na

q

n = 1 s -100dn

n!

aq

n = 1s -1dn

ln nn - ln na

q

n = 1s -1dn

nn + 1

52. The limit L of an alternating series that satisfies the conditions ofTheorem 14 lies between the values of any two consecutive par-tial sums. This suggests using the average

to estimate L. Compute

as an approximation to the sum of the alternating harmonic series.The exact sum is

53. The sign of the remainder of an alternating series that satisfiesthe conditions of Theorem 14 Prove the assertion in Theorem15 that whenever an alternating series satisfying the conditions ofTheorem 14 is approximated with one of its partial sums, then theremainder (sum of the unused terms) has the same sign as the firstunused term. (Hint: Group the remainder’s terms in consecutivepairs.)

54. Show that the sum of the first 2n terms of the series

is the same as the sum of the first n terms of the series

Do these series converge? What is the sum of the first terms of the first series? If the series converge, what is their sum?

55. Show that if diverges, then diverges.

56. Show that if converges absolutely, then

57. Show that if and both converge absolutely, thenso does

a. b.

c. (k any number)

58. Show by example that may diverge even if and both converge.

59. In Example 6, suppose the goal is to arrange the terms to get anew series that converges to Start the new arrangementwith the first negative term, which is Whenever you have asum that is less than or equal to start introducing positiveterms, taken in order, until the new total is greater than Then add negative terms until the total is less than or equal to

again. Continue this process until your partial sums have-1>2-1>2.

-1>2,-1>2.

-1>2.

gq

n=1 bn

gq

n=1 angq

n=1 an bn

aq

n = 1 kan

aq

n = 1san - bnda

q

n = 1san + bnd

gq

n=1 bngq

n=1 an

` aq

n = 1an ` … a

q

n = 1 ƒ an ƒ .

gq

n=1 an

gq

n=1 ƒ an ƒgq

n=1 an

2n + 1

11 # 2

+

12 # 3

+

13 # 4

+

14 # 5

+

15 # 6

+Á .

1 -

12

+

12

-

13

+

13

-

14

+

14

-

15

+

15

-

16

+Á

ln 2 = 0.6931. Á

s20 +

12

# 121

sn + sn + 1

2= sn +

12

s -1dn + 2an + 1

It can be shown that the sum is ln 2.

As you will see in Section 11.7, thesum is ln (1.01).

As you will see in Section 11.9, thesum is cos 1, the cosine of 1 radian.

As you will see in Section 11.9,the sum is e-1 .

T

T

T



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been above the target at least three times and finish at or below it.If is the sum of the first n terms of your new series, plot thepoints to illustrate how the sums are behaving.

60. Outline of the proof of the Rearrangement Theorem (Theo-rem 17)

a. Let be a positive real number, let and letShow that for some index and for some

index

Since all the terms appear somewhere in thesequence there is an index such that if

then is at most a sum of terms with Therefore, if

b. The argument in part (a) shows that if converges

absolutely then converges and

Now show that because converges,

converges to

61. Unzipping absolutely convergent series

a. Show that if converges and

then converges.

b. Use the results in part (a) to show likewise that if converges and

then converges.gq

n=1 cn

cn = e0, if an Ú 0

an, if an 6 0,

gq

n=1 ƒ an ƒ

gq

n=1 bn

bn = ean, if an Ú 0

0, if an 6 0,

gq

n=1 ƒ an ƒ

gq

n=1 ƒ an ƒ .

gq

n=1 ƒ bn ƒgq

n=1 ƒ an ƒ

gq

n=1 bn = gq

n=1 an .gq

n=1 bn

gq

n=1 an

… aq

k = N1

ƒ ak ƒ + ƒ sN2 - L ƒ 6 P .

an

k = 1bk - L ` … ` a

n

k = 1bk - sN2 ` + ƒ sN2 - L ƒ

n Ú N3 ,m Ú N1 .amAgn

k=1 bk B - sN2n Ú N3 ,N3 Ú N25bn6 ,

a1, a2 , Á , aN2

aq

n = N1

ƒ an ƒ 6

P

2 and ƒ sN2 - L ƒ 6

P

2.

N2 Ú N1 ,N1sk = gk

n=1 an .L = gq

n=1 an ,P

sn, sndsn

In other words, if a series converges absolutely, its pos-itive terms form a convergent series, and so do its negativeterms. Furthermore,

because and

62. What is wrong here?:

Multiply both sides of the alternating harmonic series

by 2 to get

Collect terms with the same denominator, as the arrows indicate,to arrive at

The series on the right-hand side of this equation is the serieswe started with. Therefore, and dividing by S gives (Source: “Riemann’s Rearrangement Theorem” by StewartGalanor, Mathematics Teacher, Vol. 80, No. 8, 1987, pp. 675–681.)

63. Draw a figure similar to Figure 11.9 to illustrate the convergenceof the series in Theorem 14 when N 7 1.

2 = 1.2S = S ,

2S = 1 -

12

+

13

-

14

+

15

-

16

+Á .

23

-

12

+

25

-

13

+

27

-

14

+

29

-

15

+

211

-

16

+Á .

2S = 2 - 1 +

17

-

18

+

19

-

110

+

111

-

112

+Á

S = 1 -

12

+

13

-

14

+

15

-

16

+

cn = san - ƒ an ƒ d>2.bn = san + ƒ an ƒ d>2aq

n = 1an = a

q

n = 1bn + a

q

n = 1cn





Power Series

Now that we can test infinite series for convergence we can study the infinite polynomialsmentioned at the beginning of this chapter. We call these polynomials power series be-cause they are defined as infinite series of powers of some variable, in our case x. Likepolynomials, power series can be added, subtracted, multiplied, differentiated, and inte-grated to give new power series.

11.7



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11.7 Power Series 795

DEFINITIONS Power Series, Center, CoefficientsA power series about is a series of the form

(1)

A power series about is a series of the form

(2)

in which the center a and the coefficients are constants.c0, c1, c2, Á , cn, Á

aq

n = 0cnsx - adn

= c0 + c1sx - ad + c2sx - ad2+

Á+ cnsx - adn

+Á

x � a

aq

n = 0cn xn

= c0 + c1 x + c2 x2+

Á+ cn xn

+Á .

x � 0

Power Series and Convergence

We begin with the formal definition.

Equation (1) is the special case obtained by taking in Equation (2).

EXAMPLE 1 A Geometric Series

Taking all the coefficients to be 1 in Equation (1) gives the geometric power series

This is the geometric series with first term 1 and ratio x. It converges to forWe express this fact by writing

(3)

Up to now, we have used Equation (3) as a formula for the sum of the series on the right.We now change the focus: We think of the partial sums of the series on the right as polyno-mials that approximate the function on the left. For values of x near zero, we needtake only a few terms of the series to get a good approximation. As we move toward

or we must take more terms. Figure 11.10 shows the graphs ofand the approximating polynomials for and 8.

The function is not continuous on intervals containing where ithas a vertical asymptote. The approximations do not apply when

EXAMPLE 2 A Geometric Series

The power series

(4)

matches Equation (2) with This

is a geometric series with first term 1 and ratio The series converges forr = -

x - 22

.

a = 2, c0 = 1, c1 = -1>2, c2 = 1>4, Á , cn = s -1>2dn .

1 -12

sx - 2d +14

sx - 2d2+

Á+ a- 1

2bn

sx - 2dn+

Á

x Ú 1.x = 1,ƒsxd = 1>s1 - xd

n = 0, 1, 2 ,yn = Pnsxdƒsxd = 1>s1 - xd ,-1,x = 1,

Pnsxd

11 - x

= 1 + x + x2+

Á+ xn

+Á, -1 6 x 6 1.

ƒ x ƒ 6 1.1>s1 - xd

aq

n = 0xn

= 1 + x + x2+

Á+ xn

+Á .

a = 0





or The sum is

so

Series (4) generates useful polynomial approximations of for values of x near 2:

and so on (Figure 11.11).

EXAMPLE 3 Testing for Convergence Using the Ratio Test

For what values of x do the following power series converge?

(a)

(b)

(c)

(d) aq

n = 0n!xn

= 1 + x + 2!x2+ 3!x3

+Á

aq

n = 0 xn

n!= 1 + x +

x2

2!+

x3

3!+

Á

aq

n = 1s -1dn - 1

x2n - 1

2n - 1= x -

x3

3+

x5

5 -Á

aq

n = 1s -1dn - 1

xn

n = x -

x2

2+

x3

3-

Á

P2sxd = 1 -12

sx - 2d +14

sx - 2d2= 3 -

3x2

+

x2

4,

P1sxd = 1 -12

sx - 2d = 2 -

x2

P0sxd = 1

ƒsxd = 2>x

2x = 1 -

sx - 2d2

+

sx - 2d2

4-

Á+ a- 1

2bn

sx - 2dn+

Á, 0 6 x 6 4.

11 - r

=1

1 +

x - 22

=2x ,

0 6 x 6 4.` x - 22` 6 1


0

1

1–1

2

3

4

5

7

8

9

y2 � 1 � x � x2

y1 � 1 � x

y0 � 1

y � 11 � x

y8 � 1 � x � x2 � x3 � x4 � x5 � x6 � x7 � x8

x

y

FIGURE 11.10 The graphs of and four ofits polynomial approximations (Example 1).

ƒsxd = 1>s1 - xd

0 2

1

1

y1 � 2 �

y2 � 3 � �

y0 � 1

(2, 1) y �

3

2 3x2

x2

42x

x2x

y

FIGURE 11.11 The graphs of and its first three polynomial approxima-tions (Example 2).

ƒsxd = 2>x





Solution Apply the Ratio Test to the series where is the nth term of the seriesin question.

(a)

The series converges absolutely for It diverges if because the nthterm does not converge to zero. At we get the alternating harmonic series

which converges. At we get the negative of the harmonic series; it diverges. Series (a) con-

verges for and diverges elsewhere.

(b)

The series converges absolutely for It diverges for because the nthterm does not converge to zero. At the series becomes

which converges by the Alternating Series Theorem. It also con-verges at because it is again an alternating series that satisfies the conditionsfor convergence. The value at is the negative of the value at Series (b)converges for and diverges elsewhere.

(c)

The series converges absolutely for all x.

(d)

The series diverges for all values of x except

Example 3 illustrates how we usually test a power series for convergence, and thepossible results.

0x

x = 0.

` un + 1un` = ` sn + 1d!xn + 1

n!xn ` = sn + 1d ƒ x ƒ : q unless x = 0.

0x

` un + 1un` = ` xn + 1

sn + 1d!# n!xn ` =

ƒ x ƒ

n + 1: 0 for every x .

–1 0 1x

-1 … x … 1x = 1.x = -1

x = -11>5 - 1>7 +

Á ,1 - 1>3 +x = 1

x27 1x2

6 1.

` un + 1un` =

2n - 12n + 1

x2 : x2 .

–1 0 1x

-1 6 x … 11>3 - 1>4 -

Á ,-1 - 1>2 -x = -11 - 1>2 + 1>3 - 1>4 +

Á ,x = 1,

ƒ x ƒ 7 1ƒ x ƒ 6 1.

` un + 1un` =

nn + 1 ƒ x ƒ : ƒ x ƒ .

ung ƒ un ƒ ,

THEOREM 18 The Convergence Theorem for Power Series

If the power series converges for

then it converges absolutely for all x with If the series diverges for then it diverges for all x with ƒ x ƒ 7 ƒ d ƒ .x = d ,

ƒ x ƒ 6 ƒ c ƒ .x = c Z 0,

aq

n = 0an xn

= a0 + a1 x + a2 x2+

Á



Proof Suppose the series converges. Then Hence, there isan integer N such that for all That is,

(5)

Now take any x such that and consider

There are only a finite number of terms prior to and their sum is finite. Startingwith and beyond, the terms are less than

(6)

because of Inequality (5). But Series (6) is a geometric series with ratio whichis less than 1, since Hence Series (6) converges, so the original series convergesabsolutely. This proves the first half of the theorem.

The second half of the theorem follows from the first. If the series diverges at and converges at a value with we may take in the first half of the the-orem and conclude that the series converges absolutely at d. But the series cannot convergeabsolutely and diverge at one and the same time. Hence, if it diverges at d, it diverges forall x with

To simplify the notation, Theorem 18 deals with the convergence of series of the formFor series of the form we can replace by and apply the re-

sults to the series

The Radius of Convergence of a Power Series

The theorem we have just proved and the examples we have studied lead to the conclusionthat a power series behaves in one of three possible ways. It might convergeonly at or converge everywhere, or converge on some interval of radius R centeredat We prove this as a Corollary to Theorem 18.x = a .

x = a ,gcnsx - adn

gansx¿dn .x¿x - agansx - adngan xn .

ƒ x ƒ 7 ƒ d ƒ .

c = x0ƒ x0 ƒ 7 ƒ d ƒ ,x0

x = d

ƒ x ƒ 6 ƒ c ƒ .r = ƒ x>c ƒ ,

` xc `N

+ ` xc `N + 1

+ ` xc `N + 2

+Á

ƒ aN xNƒ

ƒ aN xNƒ ,

ƒ a0 ƒ + ƒ a1 x ƒ +Á

+ ƒ aN - 1xN - 1

ƒ + ƒ aN xNƒ + ƒ aN + 1xN + 1

ƒ +Á .

ƒ x ƒ 6 ƒ c ƒ

ƒ an ƒ 61

ƒ c ƒn for n Ú N .

n Ú N .ƒ an cnƒ 6 1

limn:q an cn= 0.gq

n=0 an cn


COROLLARY TO THEOREM 18The convergence of the series is described by one of the followingthree possibilities:

1. There is a positive number R such that the series diverges for x withbut converges absolutely for x with The series

may or may not converge at either of the endpoints and

2. The series converges absolutely for every

3. The series converges at and diverges elsewhere sR = 0d .x = a

x sR = q d .

x = a + R .x = a - R

ƒ x - a ƒ 6 R .ƒ x - a ƒ 7 R

gcnsx - adn




Proof We assume first that so that the power series is centered at 0. If the se-ries converges everywhere we are in Case 2. If it converges only at we are inCase 3. Otherwise there is a nonzero number d such that diverges. The set S ofvalues of x for which the series converges is nonempty because it contains 0and a positive number p as well. By Theorem 18, the series diverges for all x with

so for all and S is a bounded set. By the Completeness Prop-erty of the real numbers (see Appendix 4) a nonempty, bounded set has a least upperbound R. (The least upper bound is the smallest number with the property that the ele-ments satisfy ) If then so the series diverges. If

then is not an upper bound for S (because it’s smaller than the least upperbound) so there is a number such that Since the series converges and therefore the series converges by Theorem 18. This proves theCorollary for power series centered at

For a power series centered at we set and repeat the argumentwith Since when a radius R interval of convergence for cen-tered at is the same as a radius R interval of convergence for centeredat This establishes the Corollary for the general case.

R is called the radius of convergence of the power series and the interval of radius Rcentered at is called the interval of convergence. The interval of convergence maybe open, closed, or half-open, depending on the particular series. At points x with

the series converges absolutely. If the series converges for all values of x,we say its radius of convergence is infinite. If it converges only at we say its radiusof convergence is zero.

x = a ,ƒ x - a ƒ 6 R ,

x = a

x = a .gcnsx - adnx¿ = 0

gcnsx¿dnx = a ,x¿ = 0x¿ .x¿ = sx - ada Z 0,

a = 0.gcn ƒ x ƒ

ngcn bnb H S ,b 7 ƒ x ƒ .b H S

ƒ x ƒƒ x ƒ 6 R ,gcn xnx x Sƒ x ƒ 7 R Ú p ,x … R .x H S

x H S ,ƒ x ƒ … ƒ d ƒƒ x ƒ 7 ƒ d ƒ ,

gcn xngcn dn

x = 0a = 0,

How to Test a Power Series for Convergence

1. Use the Ratio Test (or nth-Root Test) to find the interval where the seriesconverges absolutely. Ordinarily, this is an open interval

2. If the interval of absolute convergence is finite, test for convergence or diver-gence at each endpoint, as in Examples 3a and b. Use a Comparison Test, theIntegral Test, or the Alternating Series Test.

3. If the interval of absolute convergence is the seriesdiverges for (it does not even converge conditionally), becausethe nth term does not approach zero for those values of x.

ƒ x - a ƒ 7 Ra - R 6 x 6 a + R ,

ƒ x - a ƒ 6 R or a - R 6 x 6 a + R .

Term-by-Term Differentiation

A theorem from advanced calculus says that a power series can be differentiated term byterm at each interior point of its interval of convergence.



EXAMPLE 4 Applying Term-by-Term Differentiation

Find series for and if

Solution

CAUTION Term-by-term differentiation might not work for other kinds of series. For ex-ample, the trigonometric series

converges for all x. But if we differentiate term by term we get the series

which diverges for all x. This is not a power series, since it is not a sum of positive integerpowers of x.

aq

n = 1 n!cos sn!xd

n2 ,

aq

n = 1 sin sn!xd

n2

= aq

n = 2nsn - 1dxn - 2, -1 6 x 6 1

ƒ–sxd =2

s1 - xd3 = 2 + 6x + 12x2+

Á+ nsn - 1dxn - 2

+Á

= aq

n = 1nxn - 1, -1 6 x 6 1

ƒ¿sxd =1

s1 - xd2 = 1 + 2x + 3x2+ 4x3

+Á

+ nxn - 1+

Á

= aq

n = 0xn, -1 6 x 6 1

ƒsxd =1

1 - x= 1 + x + x2

+ x3+ x4

+Á

+ xn+

Á

ƒ–sxdƒ¿sxd


THEOREM 19 The Term-by-Term Differentiation TheoremIf converges for for some it definesa function ƒ:

Such a function ƒ has derivatives of all orders inside the interval of convergence.We can obtain the derivatives by differentiating the original series term by term:

and so on. Each of these derived series converges at every interior point of the in-terval of convergence of the original series.

ƒ–sxd = aq

n = 2nsn - 1dcnsx - adn - 2 ,

ƒ¿sxd = aq

n = 1ncnsx - adn - 1

ƒsxd = aq

n = 0cnsx - adn, a - R 6 x 6 a + R .

R 7 0,a - R 6 x 6 a + Rgcnsx - adn




EXAMPLE 5 A Series for

Identify the function

Solution We differentiate the original series term by term and get

This is a geometric series with first term 1 and ratio so

We can now integrate to get

The series for ƒ(x) is zero when so Hence

(7)

In Section 11.10, we will see that the series also converges to at x = ;1.tan-1 x

ƒsxd = x -

x3

3+

x5

5 -

x7

7 +Á

= tan-1 x, -1 6 x 6 1.

C = 0.x = 0,

Lƒ¿sxd dx = L dx

1 + x2 = tan-1 x + C .

ƒ¿sxd = 1>s1 + x2d

ƒ¿sxd =1

1 - s -x2d=

11 + x2 .

-x2 ,

ƒ¿sxd = 1 - x2+ x4

- x6+

Á, -1 6 x 6 1.

ƒsxd = x -

x3

3+

x5

5 -Á, -1 … x … 1.

tan-1 x, -1 … x … 1

THEOREM 20 The Term-by-Term Integration TheoremSuppose that

converges for Then

converges for and

for a - R 6 x 6 a + R .

Lƒsxd dx = aq

n = 0cn

sx - adn + 1

n + 1+ C

a - R 6 x 6 a + R

aq

n =0cn

(x - a)n+1

n + 1

a - R 6 x 6 a + R sR 7 0d .

ƒsxd = aq

n = 0cnsx - adn

Term-by-Term Integration

Another advanced calculus theorem states that a power series can be integrated term byterm throughout its interval of convergence.




USING TECHNOLOGY Study of Series

Series are in many ways analogous to integrals. Just as the number of functions with ex-plicit antiderivatives in terms of elementary functions is small compared to the numberof integrable functions, the number of power series in x that agree with explicit elemen-tary functions on x-intervals is small compared to the number of power series that con-verge on some x-interval. Graphing utilities can aid in the study of such series in muchthe same way that numerical integration aids in the study of definite integrals. The abilityto study power series at particular values of x is built into most Computer Algebra Sys-tems.

If a series converges rapidly enough, CAS exploration might give us an idea of thesum. For instance, in calculating the early partial sums of the series (Section 11.4, Example 2b), Maple returns for Thissuggests that the sum of the series is 1.6066 95152 to 10 digits. Indeed,

The remainder after 200 terms is negligible.However, CAS and calculator exploration cannot do much for us if the series con-

verges or diverges very slowly, and indeed can be downright misleading. For example,try calculating the partial sums of the series The terms are tiny incomparison to the numbers we normally work with and the partial sums, even for hun-dreds of terms, are miniscule. We might well be fooled into thinking that the series con-verges. In fact, it diverges, as we can see by writing it as a constanttimes the harmonic series.

We will know better how to interpret numerical results after studying error estimatesin Section 11.9.

s1>1010dgq

k=1 s1>kd ,

gq

k=1 [1>s1010kd] .

aq

k = 201

12k

- 1= a

q

k = 201

12k - 1s2 - s1>2k - 1dd

6 aq

k = 201

12k - 1 =

12199 6 1.25 * 10-60 .

31 … n … 200.Sn = 1.6066 95152gq

k=1 [1>s2k - 1d]

Notice that the original series in Example 5 converges at both endpoints of the origi-nal interval of convergence, but Theorem 20 can guarantee the convergence of the differ-entiated series only inside the interval.

EXAMPLE 6 A Series for

The series

converges on the open interval Therefore,

It can also be shown that the series converges at to the number ln 2, but that was notguaranteed by the theorem.

x = 1

= x -

x2

2+

x3

3-

x4

4+

Á, -1 6 x 6 1.

ln s1 + xd = Lx

0

11 + t

dt = t -

t2

2+

t3

3-

t4

4+

Á d0

x

-1 6 t 6 1.

11 + t

= 1 - t + t2- t3

+Á

ln s1 + xd, -1 6 x … 1

Theorem 20




EXAMPLE 7 Multiply the geometric series

by itself to get a power series for for

Solution Let

and

Then, by the Series Multiplication Theorem,

is the series for The series all converge absolutely for Notice that Example 4 gives the same answer because

ddx

a 11 - x

b =1

s1 - xd2 .

ƒ x ƒ 6 1.1>s1 - xd2 .

= 1 + 2x + 3x2+ 4x3

+Á

+ sn + 1dxn+

Á

Asxd # Bsxd = aq

n = 0cn xn

= aq

n = 0sn + 1dxn

n + 1 ones('''')''''*

= 1 + 1 +Á

+ 1 = n + 1.

n + 1 terms(''''''''''')''''''''''''*

cn = a0 bn + a1 bn - 1 +Á

+ ak bn - k +Á

+ an b0

Bsxd = aq

n = 0bn xn

= 1 + x + x2+

Á+ xn

+Á

= 1>s1 - xd

Asxd = aq

n = 0an xn

= 1 + x + x2+

Á+ xn

+Á

= 1>s1 - xd

ƒ x ƒ 6 1.1>s1 - xd2 ,

aq

n = 0xn

= 1 + x + x2+

Á+ xn

+Á

=1

1 - x , for ƒ x ƒ 6 1,

THEOREM 21 The Series Multiplication Theorem for Power SeriesIf and converge absolutely for and

then converges absolutely to A(x)B(x) for

aaq

n = 0an xnb # aa

q

n = 0bn xnb = a

q

n = 0cn xn .

ƒ x ƒ 6 R :gq

n=0 cn xn

cn = a0 bn + a1 bn - 1 + a2 bn - 2 +Á

+ an - 1b1 + an b0 = an

k = 0ak bn - k ,

ƒ x ƒ 6 R ,Bsxd = gq

n=0 bn xnAsxd = gq

n=0 an xn

Multiplication of Power Series

Another theorem from advanced calculus states that absolutely converging power seriescan be multiplied the way we multiply polynomials. We omit the proof.




EXERCISES 11.7

Intervals of ConvergenceIn Exercises 1–32, (a) find the series’ radius and interval of conver-gence. For what values of x does the series converge (b) absolutely,(c) conditionally?

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27.

28.

29. 30.

31. 32. aq

n = 0 Ax - 22 B2n + 1

2naq

n = 1 sx + pdn2n

aq

n = 1 s3x + 1dn + 1

2n + 2aq

n = 1 s4x - 5d2n + 1

n3>2

aq

n = 2

xn

n ln n

aq

n = 2

xn

nsln nd2

aq

n = 0s -2dnsn + 1dsx - 1dna

q

n = 1 s -1dn + 1sx + 2dn

n2n

aq

n = 0n!sx - 4dna

q

n = 1nnxn

aq

n = 1sln ndxna

q

n = 1 a1 +

1n b

n

xn

aq

n = 12n ns2x + 5dna

q

n = 0 2nxn

3n

aq

n = 0

nxn

4nsn2+ 1da

q

n = 0 nsx + 3dn

5n

aq

n = 0

s -1dnxn2n2+ 3

aq

n = 0

xn2n2+ 3

aq

n = 0 s2x + 3d2n + 1

n!aq

n = 0 x2n + 1

n!

aq

n = 0 3nxn

n!aq

n = 0 s -1dnxn

n!

aq

n = 1 sx - 1dn2n

aq

n = 1

xn

n2n 3n

aq

n = 1 s -1dnsx + 2dn

naq

n = 0

nxn

n + 2

aq

n = 0s2xdna

q

n = 0 sx - 2dn

10n

aq

n = 1 s3x - 2dn

naq

n = 0s -1dns4x + 1dn

aq

n = 0sx + 5dna

q

n = 0xn

In Exercises 33–38, find the series’ interval of convergence and,within this interval, the sum of the series as a function of x.

33. 34.

35. 36.

37. 38.

Theory and Examples39. For what values of x does the series

converge? What is its sum? What series do you get if you differ-entiate the given series term by term? For what values of x doesthe new series converge? What is its sum?

40. If you integrate the series in Exercise 39 term by term, what newseries do you get? For what values of x does the new series con-verge, and what is another name for its sum?

41. The series

converges to sin x for all x.

a. Find the first six terms of a series for cos x. For what valuesof x should the series converge?

b. By replacing x by 2x in the series for sin x, find a series thatconverges to sin 2x for all x.

c. Using the result in part (a) and series multiplication, calculatethe first six terms of a series for 2 sin x cos x. Compare youranswer with the answer in part (b).

42. The series

converges to for all x.

a. Find a series for Do you get the series for Explain your answer.

b. Find a series for Do you get the series for Explain your answer.

c. Replace x by in the series for to find a series thatconverges to for all x. Then multiply the series for and

to find the first six terms of a series for e-x # ex .e-xexe-x

ex-x

ex ?1ex dx .

ex ?sd>dxdex .

ex

ex= 1 + x +

x2

2!+

x3

3!+

x4

4!+

x5

5!+

Á

sin x = x -

x3

3!+

x5

5!-

x7

7!+

x9

9!-

x11

11!+

Á

1 -

12

sx - 3d +

14

sx - 3d2+

Á+ a- 1

2bn

sx - 3dn+

Á

aq

n = 0 ax2

- 12bn

aq

n = 0 ax2

+ 13bn

aq

n = 0sln xdna

q

n = 0 a2x

2- 1bn

aq

n = 0 sx + 1d2n

9naq

n = 0 sx - 1d2n

4n

Get the information you need aboutfrom Section 11.3,

Exercise 39.a1>(n(ln n)2)

Get the information you need aboutfrom Section 11.3,

Exercise 38.a1>(n ln n)



tcu1107a.html

tcu1107a.html

tcu1107b.html

tcu1107b.html

805

43. The series

converges to tan x for

a. Find the first five terms of the series for For whatvalues of x should the series converge?

b. Find the first five terms of the series for For whatvalues of x should this series converge?

c. Check your result in part (b) by squaring the series given forsec x in Exercise 44.

44. The series

converges to sec x for

a. Find the first five terms of a power series for the functionFor what values of x should the series

converge?

b. Find the first four terms of a series for sec x tan x. For whatvalues of x should the series converge?

ln ƒ sec x + tan x ƒ .

-p>2 6 x 6 p>2.

sec x = 1 +

x2

2+

524

x4+

61720

x6+

2778064

x8+

Á

sec2 x .

ln ƒ sec x ƒ .

-p>2 6 x 6 p>2.

tan x = x +

x3

3+

2x5

15+

17x7

315+

62x9

2835+

Á

c. Check your result in part (b) by multiplying the series for sec xby the series given for tan x in Exercise 43.

45. Uniqueness of convergent power series

a. Show that if two power series and areconvergent and equal for all values of x in an open interval

then for every n. (Hint: LetDifferentiate term by term

to show that and both equal )

b. Show that if for all x in an open intervalthen for every n.

46. The sum of the series To find the sum of this se-ries, express as a geometric series, differentiate bothsides of the resulting equation with respect to x, multiply bothsides of the result by x, differentiate again, multiply by x again,and set x equal to 1 2. What do you get? (Source: David E.Dobbs’ letter to the editor, Illinois Mathematics Teacher, Vol. 33,Issue 4, 1982, p. 27.)

47. Convergence at endpoints Show by examples that the conver-gence of a power series at an endpoint of its interval of conver-gence may be either conditional or absolute.

48. Make up a power series whose interval of convergence is

a. b. c. (1, 5).s -2, 0ds -3, 3d

>

1>s1 - xdgq

n=0 sn2>2ndan = 0s -c, cd ,gq

n=0 an xn= 0

f snds0d>sn!d .bnan

ƒsxd = gq

n=0 an xn= gq

n=0 bn xn .an = bns -c, cd ,

gq

n=0 bn xngq

n=0 an xn


11.7 Power Series


11.8 Taylor and Maclaurin Series 805

Taylor and Maclaurin Series

This section shows how functions that are infinitely differentiable generate power seriescalled Taylor series. In many cases, these series can provide useful polynomial approxima-tions of the generating functions.

Series Representations

We know from Theorem 19 that within its interval of convergence the sum of a powerseries is a continuous function with derivatives of all orders. But what about the other wayaround? If a function ƒ(x) has derivatives of all orders on an interval I, can it be expressedas a power series on I? And if it can, what will its coefficients be?

We can answer the last question readily if we assume that ƒ(x) is the sum of a powerseries

with a positive radius of convergence. By repeated term-by-term differentiation within theinterval of convergence I we obtain

ƒ‡sxd = 1 # 2 # 3a3 + 2 # 3 # 4a4sx - ad + 3 # 4 # 5a5sx - ad2+

Á ,

ƒ–sxd = 1 # 2a2 + 2 # 3a3sx - ad + 3 # 4a4sx - ad2+

Á

ƒ¿sxd = a1 + 2a2sx - ad + 3a3sx - ad2+

Á+ nansx - adn - 1

+Á

= a0 + a1sx - ad + a2sx - ad2+

Á+ ansx - adn

+Á

ƒsxd = aq

n = 0ansx - adn

11.8




with the nth derivative, for all n, being

Since these equations all hold at we have

and, in general,

These formulas reveal a pattern in the coefficients of any power series that converges to the values of ƒ on I (“represents ƒ on I”). If there is such a series (still anopen question), then there is only one such series and its nth coefficient is

If ƒ has a series representation, then the series must be

(1)

But if we start with an arbitrary function ƒ that is infinitely differentiable on an interval Icentered at and use it to generate the series in Equation (1), will the series then con-verge to ƒ(x) at each x in the interior of I? The answer is maybe—for some functions it willbut for other functions it will not, as we will see.

Taylor and Maclaurin Series

x = a

+Á

+

ƒsndsadn!

sx - adn+

Á .

ƒsxd = ƒsad + ƒ¿sadsx - ad +

ƒ–sad2!

sx - ad2

an =

ƒsndsadn!

.

gq

n=0 ansx - adn

ƒsndsad = n!an .

ƒ¿sad = a1,

ƒ–sad = 1 # 2a2,

ƒ‡sad = 1 # 2 # 3a3,

x = a ,

f sndsxd = n!an + a sum of terms with sx - ad as a factor .

DEFINITIONS Taylor Series, Maclaurin SeriesLet ƒ be a function with derivatives of all orders throughout some interval con-taining a as an interior point. Then the Taylor series generated by ƒ at is

The Maclaurin series generated by ƒ is

the Taylor series generated by ƒ at x = 0.

aq

k = 0 ƒskds0d

k! xk

= ƒs0d + ƒ¿s0dx +

ƒ–s0d2!

x2+

Á+

ƒsnds0dn!

xn+

Á ,

+Á

+

ƒsndsadn!

sx - adn+

Á .

aq

k = 0 ƒskdsad

k! sx - adk

= ƒsad + ƒ¿sadsx - ad +

ƒ–sad2!

sx - ad2

x = a

HISTORICAL BIOGRAPHIES

Brook Taylor(1685–1731)

Colin Maclaurin(1698–1746)

The Maclaurin series generated by ƒ is often just called the Taylor series of ƒ.



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11.8 Taylor and Maclaurin Series 807

EXAMPLE 1 Finding a Taylor Series

Find the Taylor series generated by at Where, if anywhere, does theseries converge to 1 x?

Solution We need to find Taking derivatives we get

The Taylor series is

This is a geometric series with first term 1 2 and ratio It converges ab-solutely for and its sum is

In this example the Taylor series generated by at converges to 1 x foror

Taylor Polynomials

The linearization of a differentiable function ƒ at a point a is the polynomial of degree onegiven by

In Section 3.8 we used this linearization to approximate ƒ(x) at values of x near a. If ƒ hasderivatives of higher order at a, then it has higher-order polynomial approximations aswell, one for each available derivative. These polynomials are called the Taylor polyno-mials of ƒ.

P1sxd = ƒsad + ƒ¿sadsx - ad .

0 6 x 6 4.ƒ x - 2 ƒ 6 2>a = 2ƒsxd = 1>x

1>21 + sx - 2d>2 =

12 + sx - 2d

=1x .

ƒ x - 2 ƒ 6 2r = -sx - 2d>2.>

=12

-

sx - 2d22 +

sx - 2d2

23 -Á

+ s -1dn sx - 2dn

2n + 1 +Á .

ƒs2d + ƒ¿s2dsx - 2d +

ƒ–s2d2!

sx - 2d2+

Á+

ƒsnds2dn!

sx - 2dn+

Á

ƒsndsxd = s -1dnn!x-sn + 1d, ƒsnds2d

n!=

s -1dn

2n + 1 .

o o

ƒ‡sxd = -3!x-4, ƒ‡s2d

3!= -

124 ,

ƒ–sxd = 2!x-3, ƒ–s2d

2!= 2-3

=123 ,

ƒ¿sxd = -x-2, ƒ¿s2d = -122 ,

ƒsxd = x-1, ƒs2d = 2-1=

12

,

ƒs2d, ƒ¿s2d, ƒ–s2d, Á .

> a = 2.ƒsxd = 1>x



We speak of a Taylor polynomial of order n rather than degree n because maybe zero. The first two Taylor polynomials of at for example, are

and The first-order Taylor polynomial has degree zero, not one.Just as the linearization of ƒ at provides the best linear approximation of ƒ in

the neighborhood of a, the higher-order Taylor polynomials provide the best polynomialapproximations of their respective degrees. (See Exercise 32.)

EXAMPLE 2 Finding Taylor Polynomials for

Find the Taylor series and the Taylor polynomials generated by at

Solution Since

we have

The Taylor series generated by ƒ at is

This is also the Maclaurin series for In Section 11.9 we will see that the series con-verges to at every x.

The Taylor polynomial of order n at is

See Figure 11.12.

EXAMPLE 3 Finding Taylor Polynomials for cos x

Find the Taylor series and Taylor polynomials generated by at x = 0.ƒsxd = cos x

Pnsxd = 1 + x +

x2

2+

Á+

xn

n! .

x = 0ex

ex .

= aq

k = 0 xk

k!.

= 1 + x +

x2

2+

Á+

xn

n!+

Á

ƒs0d + ƒ¿s0dx +

ƒ–s0d2!

x2+

Á+

ƒsnds0dn!

xn+

Á

x = 0

ƒs0d = e0= 1, ƒ¿s0d = 1, Á , ƒsnds0d = 1, . Á

ƒsxd = ex, ƒ¿sxd = ex, Á , ƒsndsxd = ex, Á ,

x = 0.ƒsxd = ex

ex

x = aP1sxd = 1.P0sxd = 1

x = 0,ƒsxd = cos xƒsndsad


DEFINITION Taylor Polynomial of Order nLet ƒ be a function with derivatives of order k for in some inter-val containing a as an interior point. Then for any integer n from 0 through N, theTaylor polynomial of order n generated by ƒ at is the polynomial

+

ƒskdsadk!

sx - adk+

Á+

ƒsndsadn!

sx - adn .

Pnsxd = ƒsad + ƒ¿sadsx - ad +

ƒ–sad2!

sx - ad2+

Á

x = a

k = 1, 2, Á , N

0.5

1.0

y � e x

0 0.5

1.5

2.0

2.5

3.0y � P3(x)

y � P2(x)

y � P1(x)

1.0

x

y

–0.5

FIGURE 11.12 The graph of and its Taylor polynomials

Notice the very close agreement near thecenter (Example 2).x = 0

P3sxd = 1 + x + sx2>2!d + sx3>3!d .

P2sxd = 1 + x + sx2>2!d P1sxd = 1 + x

ƒsxd = ex



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11.8 Taylor and Maclaurin Serie 809

Solution The cosine and its derivatives are

At the cosines are 1 and the sines are 0, so

The Taylor series generated by ƒ at 0 is

This is also the Maclaurin series for cos x. In Section 11.9, we will see that the series con-verges to cos x at every x.

Because the Taylor polynomials of orders 2n and are identical:

Figure 11.13 shows how well these polynomials approximate near Only the right-hand portions of the graphs are given because the graphs are symmetricabout the y-axis.

x = 0.ƒsxd = cos x

P2nsxd = P2n + 1sxd = 1 -

x2

2!+

x4

4!-

Á+ s -1dn

x2n

s2nd!.

2n + 1ƒs2n + 1ds0d = 0,

= aq

k = 0 s -1dkx2k

s2kd!.

= 1 + 0 # x -

x2

2!+ 0 # x3

+

x4

4!+

Á+ s -1dn

x2n

s2nd!+

Á

ƒs0d + ƒ¿s0dx +

ƒ–s0d2!

x2+

ƒ‡s0d3!

x3+

Á+

ƒsnds0dn!

xn+

Á

ƒs2nds0d = s -1dn, ƒs2n + 1ds0d = 0.

x = 0,

sin x . ƒs2n + 1dsxd = s -1dn + 1 cos x, ƒs2ndsxd = s -1dn

o o

sin x, ƒs3dsxd = -cos x, ƒ–sxd = -sin x, ƒ¿sxd = cos x, ƒsxd =

0 1

1y � cos x

2

–1

–2

2 3 4 5 6 7 9

P0P4 P8 P12 P16

P2 P6 P10 P14 P18

8x

y

FIGURE 11.13 The polynomials

converge to cos x as We can deduce the behavior of cos xarbitrarily far away solely from knowing the values of the cosineand its derivatives at (Example 3).x = 0

n : q .

P2nsxd = an

k = 0 s -1dkx2k

s2kd!



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EXAMPLE 4 A Function ƒ Whose Taylor Series Converges at Every x but Con-verges to ƒ(x) Only at

It can be shown (though not easily) that

(Figure 11.14) has derivatives of all orders at and that for all n. Thismeans that the Taylor series generated by ƒ at is

The series converges for every x (its sum is 0) but converges to ƒ(x) only at x = 0.

= 0 + 0 +Á

+ 0 +Á .

= 0 + 0 # x + 0 # x2+

Á+ 0 # xn

+Á

ƒs0d + ƒ¿s0dx +

ƒ–s0d2!

x2+

Á+

ƒsnds0dn!

xn+

Á

x = 0ƒsnds0d = 0x = 0

ƒsxd = e0, x = 0

e-1>x2

, x Z 0

x = 0


0 1 2 3 4

1

–1–2–3–4

y � e–1/x2

, x � 0

0 , x � 0

x

y

FIGURE 11.14 The graph of the continuous extension ofis so flat at the origin that all of its derivatives there

are zero (Example 4).y = e-1>x2

Two questions still remain.

1. For what values of x can we normally expect a Taylor series to converge to its generat-ing function?

2. How accurately do a function’s Taylor polynomials approximate the function on agiven interval?

The answers are provided by a theorem of Taylor in the next section.




EXERCISES 11.8

Finding Taylor PolynomialsIn Exercises 1–8, find the Taylor polynomials of orders 0, 1, 2, and 3generated by ƒ at a.

1. 2.

3. 4.

5. 6.

7. 8. ƒsxd = 2x + 4, a = 0ƒsxd = 2x, a = 4

ƒsxd = cos x, a = p>4ƒsxd = sin x, a = p>4ƒsxd = 1>sx + 2d, a = 0ƒsxd = 1>x, a = 2

ƒsxd = ln s1 + xd, a = 0ƒsxd = ln x, a = 1

Finding Taylor Series at (Maclaurin Series)Find the Maclaurin series for the functions in Exercises 9–20.

9. 10.

11. 12.

13. sin 3x 14. sin x2

11 - x

11 + x

ex>2e-x

x = 0



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811

15. 16.

17. 18.

19. 20.

Finding Taylor SeriesIn Exercises 21–28, find the Taylor series generated by ƒ at

21.

22.

23.

24.

25.

26.

27.

28.

Theory and Examples29. Use the Taylor series generated by at to show that

30. (Continuation of Exercise 29.) Find the Taylor series generated byat Compare your answer with the formula in Exercise 29.

31. Let ƒ(x) have derivatives through order n at Show that theTaylor polynomial of order n and its first n derivatives have thesame values that ƒ and its first n derivatives have at x = a .

x = a .

x = 1.ex

ex= ea c1 + sx - ad +

sx - ad2

2!+

Á d .x = aex

ƒsxd = 2x, a = 1

ƒsxd = ex, a = 2

ƒsxd = x>s1 - xd, a = 0

ƒsxd = 1>x2, a = 1

ƒsxd = 3x5- x4

+ 2x3+ x2

- 2, a = -1

ƒsxd = x4+ x2

+ 1, a = -2

ƒsxd = 2x3+ x2

+ 3x - 8, a = 1

ƒsxd = x3- 2x + 4, a = 2

x = a .

sx + 1d2x4- 2x3

- 5x + 4

sinh x =

ex- e-x

2cosh x =

ex+ e-x

2

5 cos px7 cos s -xd 32. Of all polynomials of degree n, the Taylor polynomial oforder n gives the best approximation Suppose that ƒ(x) is dif-ferentiable on an interval centered at and that

is a polynomial of degree nwith constant coefficients Let Show that if we impose on g the conditions

a.

b.

then

Thus, the Taylor polynomial is the only polynomial ofdegree less than or equal to n whose error is both zero at and negligible when compared with

Quadratic ApproximationsThe Taylor polynomial of order 2 generated by a twice-differentiablefunction ƒ(x) at is called the quadratic approximation of ƒ at

In Exercises 33–38, find the (a) linearization (Taylor polyno-mial of order 1) and (b) quadratic approximation of ƒ at

33. 34.

35. 36.

37. 38. ƒsxd = tan xƒsxd = sin x

ƒsxd = cosh xƒsxd = 1>21 - x2

ƒsxd = esin xƒsxd = ln scos xdx = 0.

x = a .x = a

sx - adn .x = a

Pnsxd

+

ƒsndsadn!

sx - adn .

gsxd = ƒsad + ƒ¿sadsx - ad +

ƒ–sad2!

sx - ad2+

Á

limx:a

Esxd

sx - adn = 0,

Esad = 0

ƒsxd - gsxd .Esxd =b0, Á , bn .b0 + b1sx - ad +

Á+ bnsx - adn

gsxd =x = a

◊

The approximation error is zero at x = a .

The error is negligible whencompared to sx - adn .


11.8 Taylor and Maclaurin Series


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11.9 Convergence of Taylor Series; Error Estimates 811

Convergence of Taylor Series; Error Estimates

This section addresses the two questions left unanswered by Section 11.8:

1. When does a Taylor series converge to its generating function?

2. How accurately do a function’s Taylor polynomials approximate the function on agiven interval?

Taylor’s Theorem

We answer these questions with the following theorem.

11.9




Taylor’s Theorem is a generalization of the Mean Value Theorem (Exercise 39). There is aproof of Taylor’s Theorem at the end of this section.

When we apply Taylor’s Theorem, we usually want to hold a fixed and treat b as an in-dependent variable. Taylor’s formula is easier to use in circumstances like these if wechange b to x. Here is a version of the theorem with this change.

THEOREM 22 Taylor’s TheoremIf ƒ and its first n derivatives are continuous on the closed intervalbetween a and b, and is differentiable on the open interval between a and b,then there exists a number c between a and b such that

+

ƒsndsadn!

sb - adn+

ƒsn + 1dscdsn + 1d!

sb - adn + 1 .

ƒsbd = ƒsad + ƒ¿sadsb - ad +

ƒ–sad2!

sb - ad2+

Á

ƒsndƒ¿, ƒ–, Á , ƒsnd

Taylor’s FormulaIf ƒ has derivatives of all orders in an open interval I containing a, then for eachpositive integer n and for each x in I,

(1)

where

(2)Rnsxd =

f sn + 1dscdsn + 1d!

sx - adn + 1 for some c between a and x .

+

ƒsndsadn!

sx - adn+ Rnsxd ,


ƒ–sad2!

sx - ad2+

Á

When we state Taylor’s theorem this way, it says that for each

The function is determined by the value of the derivative at a pointc that depends on both a and x, and which lies somewhere between them. For any value ofn we want, the equation gives both a polynomial approximation of ƒ of that order and aformula for the error involved in using that approximation over the interval I.

Equation (1) is called Taylor’s formula. The function is called the remainderof order n or the error term for the approximation of ƒ by over I. If as

for all we say that the Taylor series generated by ƒ at converges to ƒon I, and we write

Often we can estimate without knowing the value of c, as the following example illustrates.Rn

ƒsxd = aq

k = 0 ƒskdsad

k! sx - adk .

x = ax H I,n : q

Rnsxd : 0PnsxdRnsxd

ƒsn + 1dsn + 1dstRnsxd

ƒsxd = Pnsxd + Rnsxd .

x H I ,




EXAMPLE 1 The Taylor Series for Revisited

Show that the Taylor series generated by at converges to ƒ(x) for everyreal value of x.

Solution The function has derivatives of all orders throughout the interval Equations (1) and (2) with and give

and

Since is an increasing function of lies between and When x is negative,so is c, and When x is zero, and When x is positive, so is c, and

Thus,

and

Finally, because

and the series converges to for every x. Thus,

(3)

Estimating the Remainder

It is often possible to estimate as we did in Example 1. This method of estimation isso convenient that we state it as a theorem for future reference.

Rnsxd

ex= a

q

k = 0 xk

k!= 1 + x +

x2

2!+

Á+

xk

k!+

Á .

exlimn: q

Rnsxd = 0,

limn: q

xn + 1

sn + 1d!= 0 for every x ,

ƒ Rnsxd ƒ 6 ex xn + 1

sn + 1d! when x 7 0.

ƒ Rnsxd ƒ …

ƒ x ƒn + 1

sn + 1d! when x … 0,

ec6 ex .

Rnsxd = 0.ex= 1ec

6 1.ex .e0

= 1x, ecex

Rnsxd =

ec

sn + 1d! xn + 1 for some c between 0 and x .

ex= 1 + x +

x2

2!+

Á+

xn

n!+ Rnsxd

a = 0ƒsxd = exs - q , q d .I =

x = 0ƒsxd = ex

ex

Polynomial from Section11.8, Example 2

Section 11.1

THEOREM 23 The Remainder Estimation TheoremIf there is a positive constant M such that for all t between x anda, inclusive, then the remainder term in Taylor’s Theorem satisfies the in-equality

If this condition holds for every n and the other conditions of Taylor’s Theoremare satisfied by ƒ, then the series converges to ƒ(x).

ƒ Rnsxd ƒ … M ƒ x - a ƒ

n + 1

sn + 1d!.

Rnsxdƒ ƒsn + 1dstd ƒ … M



We are now ready to look at some examples of how the Remainder Estimation Theo-rem and Taylor’s Theorem can be used together to settle questions of convergence. As youwill see, they can also be used to determine the accuracy with which a function is approxi-mated by one of its Taylor polynomials.

EXAMPLE 2 The Taylor Series for sin x at

Show that the Taylor series for sin x at converges for all x.

Solution The function and its derivatives are

so

The series has only odd-powered terms and, for Taylor’s Theorem gives

All the derivatives of sin x have absolute values less than or equal to 1, so we can apply theRemainder Estimation Theorem with to obtain

Since as whatever the value of x, and theMaclaurin series for sin x converges to sin x for every x. Thus,

(4)

EXAMPLE 3 The Taylor Series for cos x at Revisited

Show that the Taylor series for cos x at converges to cos x for every value of x.

Solution We add the remainder term to the Taylor polynomial for cos x (Section 11.8,Example 3) to obtain Taylor’s formula for cos x with

cos x = 1 -

x2

2!+

x4

4!-

Á+ s -1dk

x2k

s2kd!+ R2ksxd .

n = 2k :

x = 0

x = 0

sin x = aq

k = 0 s -1dkx2k + 1

s2k + 1d!= x -

x3

3!+

x5

5!-

x7

7!+

Á .

R2k + 1sxd : 0,k : q ,s ƒ x ƒ2k + 2>s2k + 2d!d : 0

ƒ R2k + 1sxd ƒ … 1 #ƒ x ƒ

2k + 2

s2k + 2d!.

M = 1

sin x = x -

x3

3!+

x5

5!-

Á+

s -1dkx2k + 1

s2k + 1d!+ R2k + 1sxd .

n = 2k + 1,

f s2kds0d = 0 and f s2k + 1ds0d = s -1dk .

ƒ(2k)sxd = s -1dk sin x, o ƒ–sxd = ƒsxd = x = 0

x = 0


ƒ(2k + 1)sxd = s -1dk cos x ,

o

ƒ‡sxd = ƒ¿sxd = - sin x, - cos x,

sin x, cos x,




Because the derivatives of the cosine have absolute value less than or equal to 1, the Re-mainder Estimation Theorem with gives

For every value of x, as Therefore, the series converges to cos x for everyvalue of x. Thus,

(5)

EXAMPLE 4 Finding a Taylor Series by Substitution

Find the Taylor series for cos 2x at

Solution We can find the Taylor series for cos 2x by substituting 2x for x in the Taylorseries for cos x:

Equation (5) holds for implying that it holds for sothe newly created series converges for all x. Exercise 45 explains why the series is in factthe Taylor series for cos 2x.

EXAMPLE 5 Finding a Taylor Series by Multiplication

Find the Taylor series for x sin x at

Solution We can find the Taylor series for x sin x by multiplying the Taylor series forsin x (Equation 4) by x:

The new series converges for all x because the series for sin x converges for all x. Exer-cise 45 explains why the series is the Taylor series for x sin x.

Truncation Error

The Taylor series for at converges to for all x. But we still need to decide howmany terms to use to approximate to a given degree of accuracy. We get this informa-tion from the Remainder Estimation Theorem.

exexx = 0ex

= x2-

x4

3!+

x6

5!-

x8

7!+

Á .

x sin x = x ax -

x3

3!+

x5

5!-

x7

7!+

Áb

x = 0.

- q 6 2x 6 q ,- q 6 x 6 q ,

= aq

k = 0s -1dk

22kx2k

s2kd!.

= 1 -

22x2

2!+

24x4

4!-

26x6

6!+

Á

cos 2x = aq

k = 0 s -1dks2xd2k

s2kd!= 1 -

s2xd2

2!+

s2xd4

4!-

s2xd6

6!+

Á

x = 0.

cos x = aq

k = 0 s -1dkx2k

s2kd!= 1 -

x2

2!+

x4

4!-

x6

6!+

Á .

k : q .R2k : 0

ƒ R2ksxd ƒ … 1 #ƒ x ƒ

2k + 1

s2k + 1d!.

M = 1

Equation (5)with 2x for x



EXAMPLE 6 Calculate e with an error of less than

Solution We can use the result of Example 1 with to write

with

For the purposes of this example, we assume that we know that Hence, we arecertain that

because for By experiment we find that while Thus we should take

to be at least 10, or n to be at least 9. With an error of less than

EXAMPLE 7 For what values of x can we replace sin x by with an error ofmagnitude no greater than

Solution Here we can take advantage of the fact that the Taylor series for sin x is an al-ternating series for every nonzero value of x. According to the Alternating Series Estima-tion Theorem (Section 11.6), the error in truncating

after is no greater than

Therefore the error will be less than or equal to if

The Alternating Series Estimation Theorem tells us something that the RemainderEstimation Theorem does not: namely, that the estimate for sin x is an under-estimate when x is positive because then is positive.

Figure 11.15 shows the graph of sin x, along with the graphs of a number of its ap-proximating Taylor polynomials. The graph of is almost indistin-guishable from the sine curve when -1 … x … 1.

P3sxd = x - sx3>3!d

x5>120x - sx3>3!d

ƒ x ƒ5

1206 3 * 10-4 or ƒ x ƒ 6

52360 * 10-4L 0.514.

3 * 10-4

` x5

5!` =

ƒ x ƒ5

120.

sx3>3!d

sin x = x -

x3

3! +

x5

5!-

Á

3 * 10-4?x - sx3>3!d

e = 1 + 1 +12

+13!

+Á

+19!

L 2.718282.

10-6 ,sn + 1d3>10! 6 10-6 .1>9! 7 10-6 ,

0 6 c 6 1.1 6 ec6 3

1sn + 1d!

6 Rns1d 6

3sn + 1d!

e 6 3.

Rns1d = ec 1

sn + 1d!

e = 1 + 1 +12!

+Á

+1n!

+ Rns1d ,

x = 1

10-6 .


for some c between 0 and 1.

Rounded down,to be safe

----

--

4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM 6



You might wonder how the estimate given by the Remainder Estimation Theorem com-pares with the one just obtained from the Alternating Series Estimation Theorem. If wewrite

then the Remainder Estimation Theorem gives

which is not as good. But if we recognize that is the Taylor polynomial of order 4 as well as of order 3, then

and the Remainder Estimation Theorem with gives

This is what we had from the Alternating Series Estimation Theorem.

Combining Taylor Series

On the intersection of their intervals of convergence, Taylor series can be added, subtracted,and multiplied by constants, and the results are once again Taylor series. The Taylor seriesfor is the sum of the Taylor series for ƒ(x) and g(x) because the nth derivativeof is and so on. Thus we obtain the Taylor series for byadding 1 to the Taylor series for cos 2x and dividing the combined results by 2, and theTaylor series for is the term-by-term sum of the Taylor series for sin x andcos x.

sin x + cos x

s1 + cos 2xd>2f snd+ g snd ,f + g

ƒsxd + gsxd

ƒ R4 ƒ … 1 # ƒ x ƒ

5

5!=

ƒ x ƒ5

120.

M = 1

sin x = x -

x3

3!+ 0 + R4 ,

sx3/3!d + 0x4x - sx3>3!d = 0 + x + 0x2

-

ƒ R3 ƒ … 1 # ƒ x ƒ

4

4!=

ƒ x ƒ4

24,

sin x = x -

x3

3!+ R3 ,

1

y � sin x

2 3 4 8 9

P1 P5

P3 P7 P11 P15 P19

P9 P13 P17

5 6 70

1

2

–1

–2

x

y

FIGURE 11.15 The polynomials

converge to sin x as Notice how closely approximates the sine curve for (Example 7).x 6 1

P3sxdn : q .

P2n + 1sxd = an

k = 0 s -1dkx2k + 1

s2k + 1d!



Euler’s Identity

As you may recall, a complex number is a number of the form where a and b are

real numbers and If we substitute ( real) in the Taylor series for anduse the relations

and so on, to simplify the result, we obtain

This does not prove that because we have not yet defined what itmeans to raise e to an imaginary power. Rather, it says how to define to be consistentwith other things we know.

eiueiu

= cos u + i sin u

= a1 -

u2

2!+

u4

4!-

u6

6!+

Áb + i au -

u3

3!+

u5

5!-

Áb = cos u + i sin u .

eiu= 1 +

iu1!

+

i2u2

2!+

i3u3

3!+

i4u4

4!+

i5u5

5!+

i6u6

6!+

Á

i2= -1, i3

= i2i = - i, i4= i2i2

= 1, i5= i4i = i ,

exux = iui = 2-1.

a + bi ,


DEFINITION

(6)For any real number u, eiu= cos u + i sin u .

Equation (6), called Euler’s identity, enables us to define to be for anycomplex number One consequence of the identity is the equation

When written in the form this equation combines five of the most importantconstants in mathematics.

A Proof of Taylor’s Theorem

We prove Taylor’s theorem assuming The proof for is nearly the same.The Taylor polynomial

and its first n derivatives match the function ƒ and its first n derivatives at We donot disturb that matching if we add another term of the form where K is anyconstant, because such a term and its first n derivatives are all equal to zero at Thenew function

and its first n derivatives still agree with ƒ and its first n derivatives at We now choose the particular value of K that makes the curve agree with

the original curve at In symbols,

(7)ƒsbd = Pnsbd + Ksb - adn + 1, or K =

ƒsbd - Pnsbdsb - adn + 1 .

x = b .y = ƒsxdy = fnsxd

x = a .

fnsxd = Pnsxd + Ksx - adn + 1

x = a .Ksx - adn + 1 ,

x = a .

Pnsxd = ƒsad + ƒ¿sadsx - ad +

ƒ–sad2!

sx - ad2+

Á+

f sndsadn!

sx - adn

a 7 ba 6 b .

eip+ 1 = 0,

eip= -1.

a + bi .ea # ebiea + bi




With K defined by Equation (7), the function

measures the difference between the original function ƒ and the approximating function for each x in [a, b].

We now use Rolle’s Theorem (Section 4.2). First, because and bothF and are continuous on [a, b], we know that

Next, because and both and are continuous on we knowthat

Rolle’s Theorem, applied successively to implies the existence of

Finally, because is continuous on and differentiable on andRolle’s Theorem implies that there is a number in

such that

(8)

If we differentiate a total of times, we get

(9)

Equations (8) and (9) together give

(10)

Equations (7) and (10) give

This concludes the proof.

ƒsbd = Pnsbd +


sb - adn + 1 .

K =

ƒsn + 1dscdsn + 1d! for some number c = cn + 1 in sa, bd .

F sn + 1dsxd = ƒsn + 1dsxd - 0 - sn + 1d!K .

n + 1Fsxd = ƒsxd - Pnsxd - Ksx - adn + 1

F sn + 1dscn + 1d = 0.

sa, cndcn + 1F sndsad = F sndscnd = 0,sa, cnd ,[a, cn]F snd

cn in sa, cn - 1d such that F sndscnd = 0.

o

c4 in sa, c3d such that F s4dsc4d = 0,

c3 in sa, c2d such that F‡sc3d = 0,

F–, F‡, Á , F sn - 1d

F–sc2d = 0 for some c2 in sa, c1d .

[a, c1] ,F–F¿F¿sad = F¿sc1d = 0

F¿sc1d = 0 for some c1 in sa, bd .

F¿

Fsad = Fsbd = 0

fn

Fsxd = ƒsxd - fnsxd




EXERCISES 11.9

Taylor Series by SubstitutionUse substitution (as in Example 4) to find the Taylor series at ofthe functions in Exercises 1–6.

1. 2. 3.

4. 5. 6. cos Ax3>2>22 Bcos 2x + 1sin apx2b

5 sin s -xde-x>2e-5x

x = 0

More Taylor SeriesFind Taylor series at for the functions in Exercises 7–18.

7. 8. 9.

10. 11. 12. x2 cos sx2dx cos pxsin x - x +

x3

3!

x2

2- 1 + cos xx2 sin xxex

x = 0



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13. (Hint: )

14. 15. 16.

17. 18.

Error Estimates19. For approximately what values of x can you replace sin x by

with an error of magnitude no greater thanGive reasons for your answer.

20. If cos x is replaced by and what estimatecan be made of the error? Does tend to be too large,or too small? Give reasons for your answer.

21. How close is the approximation when Forwhich of these values of x is

22. The estimate is used when x is small. Esti-mate the error when

23. The approximation is used when x issmall. Use the Remainder Estimation Theorem to estimate theerror when

24. (Continuation of Exercise 23.) When the series for is analternating series. Use the Alternating Series Estimation Theoremto estimate the error that results from replacing by

when Compare your estimatewith the one you obtained in Exercise 23.

25. Estimate the error in the approximation when (Hint: Use not )

26. When show that may be replaced by with an error of magnitude no greater than 0.6% of h. Use

27. For what positive values of x can you replace by x withan error of magnitude no greater than 1% of the value of x?

28. You plan to estimate by evaluating the Maclaurin series forat Use the Alternating Series Estimation Theorem

to determine how many terms of the series you would have to addto be sure the estimate is good to two decimal places.

29. a. Use the Taylor series for sin x and the Alternating Series Esti-mation Theorem to show that

b. Graph together with the functionsand for Comment on

the relationships among the graphs.

30. a. Use the Taylor series for cos x and the Alternating Series Esti-mation Theorem to show that

(This is the inequality in Section 2.2, Exercise 52.)

12

-

x2

246

1 - cos x

x2 6

12

, x Z 0.

-5 … x … 5.y = 1y = 1 - sx2>6dƒsxd = ssin xd>x

1 -

x2

66

sin xx 6 1, x Z 0.

x = 1.tan-1 xp>4

ln s1 + xde0.01

= 1.01 .

1 + heh0 … h … 0.01 ,

R3 .R4 ,ƒ x ƒ 6 0.5 .sinh x = x + sx3>3!d

-0.1 6 x 6 0.1 + x + sx2>2dex

exx 6 0,

ƒ x ƒ 6 0.1 .

ex= 1 + x + sx2>2d

ƒ x ƒ 6 0.01 .21 + x = 1 + sx>2d

x 6 sin x?ƒ x ƒ 6 10-3 ?sin x = x

1 - sx2>2dƒ x ƒ 6 0.5 ,1 - sx2>2d

5 * 10-4 ?x - sx3>6d

2s1 - xd3

1s1 - xd2

x ln s1 + 2xdx2

1 - 2xsin2 x

cos2 x = s1 + cos 2xd>2.cos2 x b. Graph together withand for

Comment on the relationships among the graphs.

Finding and Identifying Maclaurin SeriesRecall that the Maclaurin series is just another name for the Taylorseries at Each of the series in Exercises 31–34 is the value ofthe Maclaurin series of a function ƒ(x) at some point. What functionand what point? What is the sum of the series?

31.

32.

33.

34.

35. Multiply the Maclaurin series for and sin x together to find thefirst five nonzero terms of the Maclaurin series for

36. Multiply the Maclaurin series for and cos x together to find thefirst five nonzero terms of the Maclaurin series for

37. Use the identity to obtain the Maclaurinseries for Then differentiate this series to obtain theMaclaurin series for 2 sin x cos x. Check that this is the series forsin 2x.

38. (Continuation of Exercise 37.) Use the identity to obtain a power series for

Theory and Examples39. Taylor’s Theorem and the Mean Value Theorem Explain how

the Mean Value Theorem (Section 4.2, Theorem 4) is a specialcase of Taylor’s Theorem.

40. Linearizations at inflection points Show that if the graph of atwice-differentiable function ƒ(x) has an inflection point at

then the linearization of ƒ at is also the quadraticapproximation of ƒ at This explains why tangent lines fitso well at inflection points.

41. The (second) second derivative test Use the equation

to establish the following test.Let ƒ have continuous first and second derivatives and sup-

pose that Then

a. ƒ has a local maximum at a if throughout an intervalwhose interior contains a;

b. ƒ has a local minimum at a if throughout an intervalwhose interior contains a.

ƒ– Ú 0

ƒ– … 0

ƒ¿sad = 0.


ƒ–sc2d2

sx - ad2

x = a .x = ax = a ,

cos2 x .cos 2x + sin2 xcos2 x =

sin2 x .sin2 x = s1 - cos 2xd>2

ex cos x .ex

ex sin x .ex

p -

p2

2+

p3

3-

Á+ s -1dk - 1

pk

k+

Á

p

3-

p3

33 # 3+

p5

35 # 5-

Á+

s -1dkp2k + 1

32k + 1s2k + 1d+

Á

1 -

p2

42 # 2!+

p4

44 # 4!-

Á+

s -1dkspd2k

42k # s2k!d+

Á

s0.1d -

s0.1d3

3!+

s0.1d5

5!-

Á+

s -1dks0.1d2k + 1

s2k + 1d!+

Á

x = 0.

-9 … x … 9.y = 1>2y = s1>2d - sx2>24dƒsxd = s1 - cos xd>x2


T

T



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42. A cubic approximation Use Taylor’s formula with andto find the standard cubic approximation of

at Give an upper bound for the magnitude ofthe error in the approximation when

43. a. Use Taylor’s formula with to find the quadratic approxi-mation of at (k a constant).

b. If for approximately what values of x in the interval[0, 1] will the error in the quadratic approximation be lessthan 1 100?

44. Improving approximations to

a. Let P be an approximation of accurate to n decimals. Showthat gives an approximation correct to 3n decimals.(Hint: Let )

b. Try it with a calculator.

45. The Taylor series generated by isA function defined by a power series

with a radius of convergence has a Taylor series that con-verges to the function at every point of Show this byshowing that the Taylor series generated by isthe series itself.

An immediate consequence of this is that series like

and

obtained by multiplying Taylor series by powers of x, as well asseries obtained by integration and differentiation of convergentpower series, are themselves the Taylor series generated by thefunctions they represent.

46. Taylor series for even functions and odd functions (Continua-tion of Section 11.7, Exercise 45.) Suppose that converges for all x in an open interval Show that

a. If ƒ is even, then i.e., the Taylorseries for ƒ at contains only even powers of x.

b. If ƒ is odd, then i.e., the Taylorseries for ƒ at contains only odd powers of x.

47. Taylor polynomials of periodic functions

a. Show that every continuous periodic function is bounded in magnitude by showing that

there exists a positive constant M such that forall x.

b. Show that the graph of every Taylor polynomial of positivedegree generated by must eventually move awayfrom the graph of cos x as increases. You can see this inFigure 11.13. The Taylor polynomials of sin x behave in asimilar way (Figure 11.15).

ƒ x ƒ

ƒsxd = cos x

ƒ ƒsxd ƒ … M

- q 6 x 6 q ,ƒsxd,

x = 0a0 = a2 = a4 =

Á= 0,

x = 0a1 = a3 = a5 =

Á= 0,

s -c, cd .ƒsxd = gq

n=0 an xn

x2ex= x2

+ x3+

x4

2!+

x5

3!+

Á ,

x sin x = x2-

x4

3!+

x6

5!-

x8

7!+

Á

gq

n=0 an xnƒsxd = gq

n=0 an xns -c, cd .

c 7 0gq

n=0 an xngq

n=0 an xnƒsxd = gq

n=0 an xn

P = p + x .P + sin P

p

P

>k = 3,

x = 0ƒsxd = s1 + xdkn = 2

ƒ x ƒ … 0.1 .x = 0.1>s1 - xd

ƒsxd =n = 3a = 0 48. a. Graph the curves and

together with the line

b. Use a Taylor series to explain what you see. What is

Euler’s Identity49. Use Equation (6) to write the following powers of e in the form

a. b. c.

50. Use Equation (6) to show that

51. Establish the equations in Exercise 50 by combining the formalTaylor series for and

52. Show that

a. b.

53. By multiplying the Taylor series for and sin x, find the termsthrough of the Taylor series for This series is the imag-inary part of the series for

Use this fact to check your answer. For what values of x shouldthe series for converge?

54. When a and b are real, we define with the equation

Differentiate the right-hand side of this equation to show that

Thus the familiar rule holds for k complex aswell as real.

55. Use the definition of to show that for any real numbers and

a. b.

56. Two complex numbers and are equal if and only ifand Use this fact to evaluate

from

where is a complex constant of integration.C = C1 + iC2

Le sa + ibdx dx =

a - ib

a2+ b2 e sa + ibdx

+ C ,

Le ax cos bx dx and Le ax sin bx dx

b = d .a = cc + ida + ib

e-iu= 1>eiu .eiu1eiu2

= eisu1 +u2d,

u2 ,u, u1 ,eiu

sd>dxde kx= ke kx

ddx

e sa + ibdx= sa + ibde sa + ibdx .

e sa + ibdx= eax # eibx

= eaxscos bx + i sin bxd .

e sa + ibdx

ex sin x

ex # eix= e s1 + idx .

ex sin x .x5ex

sinh iu = i sin u .cosh iu = cos u ,

e-iu .eiu

cos u =

eiu+ e-iu

2 and sin u =

eiu- e-iu

2i.

e-ip>2eip>4e-ip

a + bi .

limx:0

x - tan-1 x

x3 ?

y = 1>3.y = sx - tan-1 xd>x3y = s1>3d - sx2d>5

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COMPUTER EXPLORATIONSLinear, Quadratic, and Cubic ApproximationsTaylor’s formula with and gives the linearization of afunction at With and we obtain the standardquadratic and cubic approximations. In these exercises we explore theerrors associated with these approximations. We seek answers to twoquestions:

a. For what values of x can the function be replaced by eachapproximation with an error less than

b. What is the maximum error we could expect if we replace thefunction by each approximation over the specified interval?

Using a CAS, perform the following steps to aid in answeringquestions (a) and (b) for the functions and intervals in Exercises57–62.

Step 1: Plot the function over the specified interval.

Step 2: Find the Taylor polynomials and at

Step 3: Calculate the derivative associatedwith the remainder term for each Taylor polynomial. Plot the de-rivative as a function of c over the specified interval and estimateits maximum absolute value, M.

ƒsn + 1dscdsn + 1dst

x = 0.P3sxdP1sxd, P2sxd ,

10-2 ?

n = 3n = 2x = 0.a = 0n = 1

Step 4: Calculate the remainder for each polynomial. Us-ing the estimate M from Step 3 in place of plot over the specified interval. Then estimate the values of x thatanswer question (a).

Step 5: Compare your estimated error with the actual errorby plotting over the specified in-

terval. This will help answer question (b).

Step 6: Graph the function and its three Taylor approximationstogether. Discuss the graphs in relation to the information discov-ered in Steps 4 and 5.

57.

58.

59.

60.

61.

62. ƒsxd = ex>3 sin 2x, ƒ x ƒ … 2

ƒsxd = e-x cos 2x, ƒ x ƒ … 1

ƒsxd = scos xdssin 2xd, ƒ x ƒ … 2

ƒsxd =

x

x2+ 1

, ƒ x ƒ … 2

ƒsxd = s1 + xd3>2, -

12

… x … 2

ƒsxd =

121 + x, ƒ x ƒ …

34

EnsxdEnsxd = ƒ ƒsxd - Pnsxd ƒ

Rnsxdƒsn + 1dscd ,Rnsxd





Applications of Power Series

This section introduces the binomial series for estimating powers and roots and shows howseries are sometimes used to approximate the solution of an initial value problem, to eval-uate nonelementary integrals, and to evaluate limits that lead to indeterminate forms. Weprovide a self-contained derivation of the Taylor series for and conclude with a ref-erence table of frequently used series.

The Binomial Series for Powers and Roots

The Taylor series generated by when m is constant, is

(1)

This series, called the binomial series, converges absolutely for To derive theƒ x ƒ 6 1.

+

msm - 1dsm - 2d Á sm - k + 1dk!

xk+

Á .

1 + mx +

msm - 1d2!

x2+

msm - 1dsm - 2d3!

x3+

Á

ƒsxd = s1 + xdm ,

tan-1 x

11.10



11.10 Applications of Power Series 823

series, we first list the function and its derivatives:

We then evaluate these at and substitute into the Taylor series formula to obtainSeries (1).

If m is an integer greater than or equal to zero, the series stops after termsbecause the coefficients from on are zero.

If m is not a positive integer or zero, the series is infinite and converges for To see why, let be the term involving Then apply the Ratio Test for absolute conver-gence to see that

Our derivation of the binomial series shows only that it is generated by andconverges for The derivation does not show that the series converges to It does, but we omit the proof.

s1 + xdm .ƒ x ƒ 6 1.s1 + xdm

` uk + 1uk` = ` m - k

k + 1 x ` : ƒ x ƒ as k : q .

xk .uk

ƒ x ƒ 6 1.k = m + 1

sm + 1d

x = 0

ƒskdsxd = msm - 1dsm - 2d Á sm - k + 1ds1 + xdm - k .

o

ƒ‡sxd = msm - 1dsm - 2ds1 + xdm - 3

ƒ–sxd = msm - 1ds1 + xdm - 2

ƒ¿sxd = ms1 + xdm - 1

ƒsxd = s1 + xdm

The Binomial Series

For

where we define

and

amkb =

msm - 1dsm - 2d Á sm - k + 1dk!

for k Ú 3.

am1b = m, am

2b =

msm - 1d2!

,

s1 + xdm= 1 + a

q

k = 1 am

kb xk ,

-1 6 x 6 1,

EXAMPLE 1 Using the Binomial Series

If

and

a-1

kb =

-1s -2ds -3d Á s -1 - k + 1dk!

= s -1dk ak!k!b = s -1dk .

a-1

1b = -1, a-1

2b =

-1s -2d2!

= 1,

m = -1,



With these coefficient values and with x replaced by the binomial series formula givesthe familiar geometric series

EXAMPLE 2 Using the Binomial Series

We know from Section 3.8, Example 1, that for small. Withthe binomial series gives quadratic and higher-order approximations as well,

along with error estimates that come from the Alternating Series Estimation Theorem:

Substitution for x gives still other approximations. For example,

Power Series Solutions of Differential Equationsand Initial Value Problems

When we cannot find a relatively simple expression for the solution of an initial value prob-lem or differential equation, we try to get information about the solution in other ways. Oneway is to try to find a power series representation for the solution. If we can do so, we im-mediately have a source of polynomial approximations of the solution, which may be allthat we really need. The first example (Example 3) deals with a first-order linear differen-tial equation that could be solved with the methods of Section 9.2. The example shows how,not knowing this, we can solve the equation with power series. The second example (Exam-ple 4) deals with an equation that cannot be solved analytically by previous methods.

EXAMPLE 3 Series Solution of an Initial Value Problem

Solve the initial value problem

Solution We assume that there is a solution of the form

(2)y = a0 + a1 x + a2 x2+

Á+ an - 1x

n - 1+ an xn

+Á .

y¿ - y = x, ys0d = 1.

A1 -1x L 1 -

12x

-1

8x2 for ` 1x ` small, that is, ƒ x ƒ large.

21 - x 2L 1 -

x 2

2-

x4

8 for ƒ x 2

ƒ small

= 1 +

x2

-

x2

8+

x3

16-

5x4

128+

Á .

+

a12b a- 1

2b a- 3

2b a- 5

2b

4! x4

+Á

s1 + xd1>2= 1 +

x2

+

a12b a- 1

2b

2! x2

+

a12b a- 1

2b a- 3

2b

3! x3

m = 1>2,ƒ x ƒ21 + x L 1 + sx>2d

s1 + xd-1= 1 + a

q

k = 1s -1dkxk

= 1 - x + x2- x3

+Á

+ s -1dkxk+

Á .

-x ,





Our goal is to find values for the coefficients that make the series and its first derivative

(3)

satisfy the given differential equation and initial condition. The series is the differ-ence of the series in Equations (2) and (3):

(4)

If y is to satisfy the equation the series in Equation (4) must equal x. Sincepower series representations are unique (Exercise 45 in Section 11.7), the coefficients inEquation (4) must satisfy the equations

We can also see from Equation (2) that when so that (this being theinitial condition). Putting it all together, we have

Substituting these coefficient values into the equation for y (Equation (2)) gives

The solution of the initial value problem is As a check, we see that

and

EXAMPLE 4 Solving a Differential Equation

Find a power series solution for

(5)y– + x2y = 0.

y¿ - y = s2ex- 1d - s2ex

- 1 - xd = x .

ys0d = 2e0- 1 - 0 = 2 - 1 = 1

y = 2ex- 1 - x .

= 1 + x + 2sex- 1 - xd = 2ex

- 1 - x .

= 1 + x + 2 ax2

2!+

x3

3!+

Á+

xn

n!+

Áb('''''')''''''*

the Taylor series for ex- 1 - x

y = 1 + x + 2 # x2

2!+ 2 # x3

3!+

Á+ 2 # xn

n!+

Á

a3 =

a2

3=

23 # 2

=23!

, Á , an =

an - 1n =

2n!

, Á

a0 = 1, a1 = a0 = 1, a2 =

1 + a1

2=

1 + 12

=22

,

a0 = 1x = 0,y = a0

o

nan - an - 1 = 0

o

3a3 - a2 = 0

2a2 - a1 = 1

a1 - a0 = 0

y¿ - y = x ,

+ snan - an - 1dxn - 1+

Á .

y¿ - y = sa1 - a0d + s2a2 - a1dx + s3a3 - a2dx2+

Á

y¿ - y

y¿ = a1 + 2a2 x + 3a3 x2+

Á+ nan xn - 1

+Á

ak

Constant terms

Coefficients of x

Coefficients of x2

o

Coefficients of xn - 1

o



Solution We assume that there is a solution of the form

(6)

and find what the coefficients have to be to make the series and its second derivative

(7)

satisfy Equation (5). The series for is times the right-hand side of Equation (6):

(8)

The series for is the sum of the series in Equations (7) and (8):

(9)

Notice that the coefficient of in Equation (8) is If y and its second derivative are to satisfy Equation (5), the coefficients of the individual powers of x on the right-handside of Equation (9) must all be zero:

(10)

and for all

(11)

We can see from Equation (6) that

In other words, the first two coefficients of the series are the values of y and at Equations in (10) and the recursion formula in Equation (11) enable us to evaluate all theother coefficients in terms of and

The first two of Equations (10) give

Equation (11) shows that if then so we conclude that

and whenever or is zero. For the other coefficients we have

so that

and

a13 =

-a9

12 # 13=

-a1

4 # 5 # 8 # 9 # 12 # 13.

a5 =

-a1

5 # 4, a9 =

-a5

9 # 8=

a1

4 # 5 # 8 # 9

a12 =

-a8

11 # 12=

-a0

3 # 4 # 7 # 8 # 11 # 12

a4 =

-a0

4 # 3, a8 =

-a4

8 # 7=

a0

3 # 4 # 7 # 8

an =

-an - 4

nsn - 1d

4k + 3, ann = 4k + 2

a6 = 0, a7 = 0, a10 = 0, a11 = 0,

an = 0;an - 4 = 0,

a2 = 0, a3 = 0.

a1 .a0

x = 0.y¿

a0 = ys0d, a1 = y¿s0d .

nsn - 1dan + an - 4 = 0.

n Ú 4,

2a2 = 0, 6a3 = 0, 12a4 + a0 = 0, 20a5 + a1 = 0,

y–an - 4 .xn - 2

+Á

+ snsn - 1dan + an - 4dxn - 2+

Á .

y– + x2y = 2a2 + 6a3 x + s12a4 + a0dx2+ s20a5 + a1dx3

y– + x2y

x2y = a0 x2+ a1 x3

+ a2 x4+

Á+ an xn + 2

+Á .

x2x2y

y– = 2a2 + 3 # 2a3 x +Á

+ nsn - 1dan xn - 2+

Á

ak

y = a0 + a1 x + a2 x2+

Á+ an xn

+Á ,





The answer is best expressed as the sum of two separate series—one multiplied by theother by

Both series converge absolutely for all x, as is readily seen by the Ratio Test.

Evaluating Nonelementary Integrals

Taylor series can be used to express nonelementary integrals in terms of series. Integralslike arise in the study of the diffraction of light.

EXAMPLE 5 Express as a power series.

Solution From the series for sin x we obtain

Therefore,

EXAMPLE 6 Estimating a Definite Integral

Estimate with an error of less than 0.001.

Solution From the indefinite integral in Example 5,

The series alternates, and we find by experiment that

is the first term to be numerically less than 0.001. The sum of the preceding two terms gives

With two more terms we could estimate

with an error of less than With only one term beyond that we have

L1

0 sin x2 dx L

13

-142

+1

1320-

175600

+1

6894720L 0.310268303,

10-6 .

L1

0 sin x2 dx L 0.310268

L1

0 sin x2 dx L

13

-142

L 0.310.

111 # 5!

L 0.00076

L1

0 sin x2 dx =

13

-1

7 # 3!+

111 # 5!

-1

15 # 7!+

119 # 9!

-Á .

110 sin x2 dx

L sin x2 dx = C +

x3

3-

x7

7 # 3!+

x11

11 # 5!-

x15

15 # 7!+

x10

19 # 9!-

Á .

sin x2= x2

-

x6

3!+

x10

5!-

x14

7!+

x18

9!-

Á .

1 sin x2 dx

1 sin x2 dx

+ a1 ax -

x5

4 # 5+

x9

4 # 5 # 8 # 9-

x13

4 # 5 # 8 # 9 # 12 # 13+

Áb .

y = a0 a1 -

x4

3 # 4+

x8

3 # 4 # 7 # 8-

x12

3 # 4 # 7 # 8 # 11 # 12+

Ába1 :

a0 ,



with an error of about To guarantee this accuracy with the error formula forthe Trapezoidal Rule would require using about 8000 subintervals.

Arctangents

In Section 11.7, Example 5, we found a series for by differentiating to get

and integrating to get

However, we did not prove the term-by-term integration theorem on which this conclusiondepended. We now derive the series again by integrating both sides of the finite formula

(12)

in which the last term comes from adding the remaining terms as a geometric series withfirst term and ratio Integrating both sides of Equation (12)from to gives

where

The denominator of the integrand is greater than or equal to 1; hence

If the right side of this inequality approaches zero as Thereforeif and

(13)

We take this route instead of finding the Taylor series directly because the formulas for thehigher-order derivatives of are unmanageable. When we put in Equation (13),we get Leibniz’s formula:

Because this series converges very slowly, it is not used in approximating to many deci-mal places. The series for converges most rapidly when x is near zero. For that rea-son, people who use the series for to compute use various trigonometric identities.ptan-1 x

tan-1 xp

p4

= 1 -13

+15 -

17 +

19

-Á

+

s -1dn

2n + 1+

Á .

x = 1tan-1 x

tan-1 x = x -

x3

3+

x5

5 -

x7

7 +Á , ƒ x ƒ … 1

tan-1 x = aq

n = 0 s -1dnx2n + 1

2n + 1, ƒ x ƒ … 1.

ƒ x ƒ … 1limn:q Rnsxd = 0n : q .ƒ x ƒ … 1,

ƒ Rnsxd ƒ … Lƒ x ƒ

0t2n + 2 dt =

ƒ x ƒ2n + 3

2n + 3.

Rnsxd = Lx

0 s -1dn + 1t2n + 2

1 + t2 dt .

tan-1 x = x -

x3

3+

x5

5 -

x7

7 +Á

+ s -1dn x2n + 1

2n + 1+ Rnsxd ,

t = xt = 0r = - t2 .a = s -1dn + 1t2n + 2

11 + t2 = 1 - t2

+ t4- t6

+Á

+ s -1dnt2n+

s -1dn + 1t2n + 2

1 + t2 ,

tan-1 x = x -

x3

3+

x5

5 -

x7

7 +Á .

ddx

tan-1 x =1

1 + x2 = 1 - x2+ x4

- x6+

Á

tan-1 x

1.08 * 10-9 .





For example, if

then

and

Now Equation (13) may be used with to evaluate and with togive The sum of these results, multiplied by 4, gives

Evaluating Indeterminate Forms

We can sometimes evaluate indeterminate forms by expressing the functions involved asTaylor series.

EXAMPLE 7 Limits Using Power Series

Evaluate

Solution We represent ln x as a Taylor series in powers of This can be accom-plished by calculating the Taylor series generated by ln x at directly or by replacingx by in the series for in Section 11.7, Example 6. Either way, we obtain

from which we find that

EXAMPLE 8 Limits Using Power Series

Evaluate

limx:0

sin x - tan x

x3 .

lim x:1

ln x

x - 1= lim

x:1 a1 -

12

sx - 1d +Áb = 1.

ln x = sx - 1d -12

sx - 1d2+

Á ,

ln (1 + x)x - 1x = 1

x - 1.

limx:1

ln x

x - 1.

p .tan-1 (1>3).x = 1>3tan-1 (1>2)x = 1>2

p4

= a + b = tan-1 12

+ tan-1 13

.

tan sa + b d =

tan a + tan b

1 - tan a tan b=

12 +

13

1 -16

= 1 = tan p4

a = tan-1 12 and b = tan-1

13

,



Solution The Taylor series for sin x and tan x, to terms in are

Hence,

and

If we apply series to calculate we not only find the limit suc-cessfully but also discover an approximation formula for csc x.

EXAMPLE 9 Approximation Formula for csc x

Find

Solution

Therefore,

From the quotient on the right, we can see that if is small, then

1sin x

-1x L x # 1

3!=

x6 or csc x L

1x +

x6

.

ƒ x ƒ

limx:0

a 1sin x

-1x b = lim

x:0 §x

13!

-

x2

5!+

Á

1 -

x2

3!+

Á

¥ = 0.

=

x3 a 13!

-

x2

5!+

Ábx2 a1 -

x2

3!+

Áb= x

13!

-

x2

5!+

Á

1 -

x2

3!+

Á

.

1

sin x-

1x =

x - sin xx sin x

=

x - ax -

x3

3!+

x5

5!-

Ábx # ax -

x3

3!+

x5

5!-

Áb

limx:0

a 1sin x

-1x b .

limx:0 ss1>sin xd - s1/xdd ,

= -12

.

limx:0

sin x - tan x

x3 = limx:0

a- 12

-

x2

8-

Áb

sin x - tan x = -

x3

2-

x5

8-

Á= x3 a- 1

2-

x2

8-

Áb

sin x = x -

x3

3!+

x5

5!-

Á, tan x = x +

x3

3+

2x5

15+

Á .

x5 ,





TABLE 11.1 Frequently used Taylor series

Binomial Series

where

Note: To write the binomial series compactly, it is customary to define to be 1 and to take (even in the usually

excluded case where ), yielding If m is a positive integer, the series terminates at and the

result converges for all x.

xms1 + xdm= gq

k=0 amkbxk .x = 0

x0= 1am

0b

am1b = m, am

2b =

msm - 1d2!

, amkb =

msm - 1d Á sm - k + 1dk!

for k Ú 3.

= 1 + aq

k = 1 am

kbxk, ƒ x ƒ 6 1,

s1 + xdm= 1 + mx +

msm - 1dx2

2!+

msm - 1dsm - 2dx3

3!+

Á+

msm - 1dsm - 2d Á sm - k + 1dxk

k!+

Á

tan-1 x = x -

x3

3+

x5

5 -Á

+ s -1dn x2n + 1

2n + 1+

Á= a

q

n = 0 s -1dnx2n + 1

2n + 1, ƒ x ƒ … 1

ln 1 + x1 - x

= 2 tanh-1 x = 2 ax +

x3

3+

x5

5 +Á

+

x2n + 1

2n + 1+

Áb = 2aq

n = 0

x2n + 1

2n + 1, ƒ x ƒ 6 1

ln s1 + xd = x -

x2

2+

x3

3-

Á+ s -1dn - 1

xn

n +Á

= aq

n = 1 s -1dn - 1xn

n , -1 6 x … 1

cos x = 1 -

x2

2!+

x4

4!-

Á+ s -1dn

x2n

s2nd!+

Á= a

q

n = 0 s -1dnx2n

s2nd!, ƒ x ƒ 6 q

sin x = x -

x3

3!+

x5

5!-

Á+ s -1dn

x2n + 1

s2n + 1d!+

Á= a

q

n = 0 s -1dnx2n + 1

s2n + 1d!, ƒ x ƒ 6 q

ex= 1 + x +

x2

2!+

Á+

xn

n!+

Á= a

q

n = 0 xn

n!, ƒ x ƒ 6 q

11 + x

= 1 - x + x2-

Á+ s -xdn

+Á

= aq

n = 0s -1dnxn, ƒ x ƒ 6 1

11 - x

= 1 + x + x2+

Á+ xn

+Á

= aq

n = 0xn, ƒ x ƒ 6 1




EXERCISES 11.10

Binomial SeriesFind the first four terms of the binomial series for the functions in Ex-ercises 1–10.

1. 2. 3.

4. 5. 6. a1 -

x2b-2a1 +

x2b-2

s1 - 2xd1>2s1 - xd-1>2s1 + xd1>3s1 + xd1>2

7. 8.

9. 10.

Find the binomial series for the functions in Exercises 11–14.

11. 12. s1 + x2d3s1 + xd4

a1 -

2x b

1>3a1 +

1x b

1>2s1 + x2d-1>3s1 + x3d-1>2



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13. 14.

Initial Value ProblemsFind series solutions for the initial value problems in Exercises 15–32.

15. 16.

17. 18.

19. 20.

21. 22.

23.

24.

25. and

26. and

27. and

28. and

29. and

30. and

31. and

32. and

Approximations and Nonelementary IntegralsIn Exercises 33–36, use series to estimate the integrals’ values with anerror of magnitude less than (The answer section gives the inte-grals’ values rounded to five decimal places.)

33. 34.

35. 36.

Use series to approximate the values of the integrals in Exercises37–40 with an error of magnitude less than

37. 38.

39. 40.

41. Estimate the error if is approximated by in the

integral

42. Estimate the error if is approximated by

in the integral

In Exercises 43–46, find a polynomial that will approximate F(x)throughout the given interval with an error of magnitude less than

43. Fsxd = Lx

0 sin t2 dt, [0, 1]

10-3 .

110 cos 2t dt .

1 -

t2

+

t2

4!-

t3

6!cos 2t

110 cos t2 dt .

1 -

t4

2+

t8

4!cos t2

L1

0 1 - cos x

x2 dxL0.1

021 + x4 dx

L0.1

0e-x2

dxL0.1

0 sin x

x dx

10-8 .

L0.25

023 1 + x2 dxL

0.1

0

121 + x4 dx

L0.2

0 e-x

- 1x dxL

0.2

0 sin x2 dx

10-3 .

y s0d = 0y– - 2y¿ + y = 0, y¿s0d = 1

y s0d = ay– + x2y = x, y¿s0d = b

y s0d = ay– - x2y = 0, y¿s0d = b

y s2d = 0y– - y = -x, y¿s2d = -2

y s0d = -1y– - y = x, y¿s0d = 2

y s0d = 2y– + y = x, y¿s0d = 1

y s0d = 1y– + y = 0, y¿s0d = 0

y s0d = 0y– - y = 0, y¿s0d = 1

s1 + x2dy¿ + 2xy = 0, y s0d = 3

s1 - xdy¿ - y = 0, y s0d = 2

y¿ - x2y = 0, y s0d = 1y¿ - xy = 0, y s0d = 1

y¿ + y = 2x, y s0d = -1y¿ - y = x, y s0d = 0

y¿ + y = 1, y s0d = 2y¿ - y = 1, y s0d = 0

y¿ - 2y = 0, y s0d = 1y¿ + y = 0, y s0d = 1

a1 -

x2b4

s1 - 2xd3 44.

45. (a) [0, 0.5] (b) [0, 1]

46. (a) [0, 0.5] (b) [0, 1]

Indeterminate FormsUse series to evaluate the limits in Exercises 47–56.

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

Theory and Examples57. Replace x by in the Taylor series for to obtain a se-

ries for Then subtract this from the Taylor series forto show that for

58. How many terms of the Taylor series for should youadd to be sure of calculating ln (1.1) with an error of magnitudeless than Give reasons for your answer.

59. According to the Alternating Series Estimation Theorem, howmany terms of the Taylor series for would you have to addto be sure of finding with an error of magnitude less than

Give reasons for your answer.

60. Show that the Taylor series for diverges for

61. Estimating Pi About how many terms of the Taylor series forwould you have to use to evaluate each term on the right-

hand side of the equation

with an error of magnitude less than In contrast, the con-vergence of to is so slow that even 50 termswill not yield two-place accuracy.

62. Integrate the first three nonzero terms of the Taylor series for tan tfrom 0 to x to obtain the first three nonzero terms of the Taylorseries for ln sec x.

p2>6gq

n=1s1>n2d10-6 ?

p = 48 tan-1 118

+ 32 tan-1 157

- 20 tan-1 1

239

tan-1 x

ƒ x ƒ 7 1.ƒsxd = tan-1 x

10-3 ?p>4

tan-1 1

10-8 ?

ln s1 + xd

ln 1 + x1 - x

= 2 ax +

x3

3+

x5

5+

Á b .

ƒ x ƒ 6 1,ln s1 + xdln s1 - xd .

ln s1 + xd-x

limx:2

x2

- 4ln sx - 1d

limx:0

ln s1 + x2d1 - cos x

limx: q

sx + 1d sin 1

x + 1lim

x: q x2se-1>x2

- 1d

limy:0

tan-1 y - sin y

y3 cos ylimy:0

y - tan-1 y

y3

limu:0

sin u - u + su3>6d

u5limt:0

1 - cos t - st2>2d

t4

limx:0

ex

- e-x

xlimx:0

ex

- s1 + xdx2

Fsxd = Lx

0 ln s1 + td

t dt,

Fsxd = Lx

0 tan-1 t dt,

Fsxd = Lx

0t2e-t2

dt, [0, 1]


T

T

T



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833

63. a. Use the binomial series and the fact that

to generate the first four nonzero terms of the Taylor seriesfor What is the radius of convergence?

b. Series for Use your result in part (a) to find the firstfive nonzero terms of the Taylor series for

64. a. Series for Find the first four nonzero terms of theTaylor series for

b. Use the first three terms of the series in part (a) to estimateGive an upper bound for the magnitude of the

estimation error.

65. Obtain the Taylor series for from the series for

66. Use the Taylor series for to obtain a series for

67. Estimating Pi The English mathematician Wallis discoveredthe formula

Find to two decimal places with this formula.

68. Construct a table of natural logarithms for by using the formula in Exercise 57, but taking advan-

tage of the relationships and to reduce

the job to the calculation of relatively few logarithms by series.Start by using the following values for x in Exercise 57:

13

, 15

, 19

, 113

.

ln 10 = ln 2 + ln 5ln 8 = 3 ln 2, ln 9 = 2 ln 3 ,ln 4 = 2 ln 2, ln 6 = ln 2 + ln 3,

3, Á , 10n = 1, 2, ln n

p

p

4=

2 # 4 # 4 # 6 # 6 # 8 # Á

3 # 3 # 5 # 5 # 7 # 7 # Á.

2x>s1 - x2d2 .1>s1 - x2d

-1>s1 + xd .1>s1 + xd2

sinh-1 0.25 .

sinh-1 x = Lx

0

dt21 + t2.

sinh-1 x

cos-1 x .cos-1 x

sin-1 x .

ddx

sin-1 x = s1 - x2d-1>269. Series for Integrate the binomial series for

to show that for

70. Series for for Derive the series

by integrating the series

in the first case from x to and in the second case from to x.

71. The value of

a. Use the formula for the tangent of the difference of twoangles to show that

b. Show that

c. Find the value of gq

n=1 tan-1 2n2 .

aN

n = 1 tan-1

2n2 = tan-1 sN + 1d + tan-1 N -

p

4.

tan stan-1 sn + 1d - tan-1 sn - 1dd =

2n2

gq

n=1 tan-1s2>n2d

- qq

11 + t2 =

1t2 #

11 + s1>t2d

=

1t2 -

1t4 +

1t6 -

1t8 +

Á

tan-1 x = -

p

2-

1x +

13x3 -

15x5 +

Á, x 6 -1,

tan-1 x =

p

2-

1x +

13x3 -

15x5 +

Á, x 7 1

ƒ x ƒ 7 1tan-1 x

sin-1 x = x + aq

n = 1 1 # 3 # 5 # Á # s2n - 1d

2 # 4 # 6 # Á # s2nd

x2n + 1

2n + 1.

ƒ x ƒ 6 1,s1 - x2d-1>2sin-1 x

T

T

T


11.10 Applications of Power Series


11.11 Fourier Series 833

Fourier Series

We have seen how Taylor series can be used to approximate a function ƒ by polynomials.The Taylor polynomials give a close fit to ƒ near a particular point but the error inthe approximation can be large at points that are far away. There is another method thatoften gives good approximations on wide intervals, and often works with discontinuousfunctions for which Taylor polynomials fail. Introduced by Joseph Fourier, this method ap-proximates functions with sums of sine and cosine functions. It is well suited for analyzingperiodic functions, such as radio signals and alternating currents, for solving heat transferproblems, and for many other problems in science and engineering.

x = a ,

11.11


Jean-Baptiste Joseph Fourier

(1766–1830)



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Suppose we wish to approximate a function ƒ on the interval by a sum of sineand cosine functions,

or, in sigma notation,

(1)

We would like to choose values for the constants and thatmake a “best possible” approximation to ƒ(x). The notion of “best possible” isdefined as follows:

1. and ƒ(x) give the same value when integrated from 0 to

2. and ƒ(x) cos kx give the same value when integrated from 0 to

3. and ƒ(x) sin kx give the same value when integrated from 0 to

Altogether we impose conditions on

It is possible to choose and so that all these conditions aresatisfied, by proceeding as follows. Integrating both sides of Equation (1) from 0 to gives

since the integral over of cos kx equals zero when as does the integral ofsin kx. Only the constant term contributes to the integral of over A similarcalculation applies with each of the other terms. If we multiply both sides of Equation (1)by cos x and integrate from 0 to then we obtain

This follows from the fact that

and

L2p

0 cos px cos qx dx = L

2p

0 cos px sin mx dx = L

2p

0 sin px sin qx dx = 0

L2p

0 cos px cos px dx = p

L2p

0ƒnsxd cos x dx = pa1 .

2p

[0, 2p] .ƒna0

k Ú 1,[0, 2p]

L2p

0ƒnsxd dx = 2pa0

2pb1, b2, Á , bna0, a1, a2, Á an

L2p

0ƒnsxd sin kx dx = L

2p

0ƒsxd sin kx dx, k = 1, Á , n .

L2p

0ƒnsxd cos kx dx = L

2p

0ƒsxd cos kx dx, k = 1, Á , n ,

L2p

0ƒnsxd dx = L

2p

0ƒsxd dx ,

ƒn :2n + 1

2p sk = 1, Á , nd .ƒnsxd sin kx

2p sk = 1, Á , nd .ƒnsxd cos kx

2p .ƒnsxd

ƒnsxdb1, b2, Á , bna0, a1, a2, Á an

ƒnsxd = a0 + an

k = 1sak cos kx + bk sin kxd .

+ san cos nx + bn sin nxd

ƒnsxd = a0 + sa1 cos x + b1 sin xd + sa2 cos 2x + b2 sin 2xd +Á

[0, 2p]





whenever p, q and m are integers and p is not equal to q (Exercises 9–13). If we multiplyEquation (1) by sin x and integrate from 0 to we obtain

Proceeding in a similar fashion with

we obtain only one nonzero term each time, the term with a sine-squared or cosine-squared term. To summarize,

We chose so that the integrals on the left remain the same when is replaced by ƒ, sowe can use these equations to find and from ƒ:

(2)

(3)

(4)

The only condition needed to find these coefficients is that the integrals above must exist.If we let and use these rules to get the coefficients of an infinite series, then the re-sulting sum is called the Fourier series for ƒ(x),

(5)

EXAMPLE 1 Finding a Fourier Series Expansion

Fourier series can be used to represent some functions that cannot be represented by Taylorseries; for example, the step function ƒ shown in Figure 11.16a.

ƒsxd = e1, if 0 … x … p

2, if p 6 x … 2p .

a0 + aq

k = 1sak cos kx + bk sin kxd .

n : q

bk =1pL

2p

0ƒsxd sin kx dx, k = 1, Á , n

ak =1pL

2p

0ƒsxd cos kx dx, k = 1, Á , n

a0 =1

2pL2p

0ƒsxd dx

b1, b2, Á , bna0, a1, a2, Á an

ƒnƒn

L2p

0ƒnsxd sin kx dx = pbk, k = 1, Á , n

L2p

0ƒnsxd cos kx dx = pak, k = 1, Á , n

L2p

0ƒnsxd dx = 2pa0

cos 2x, sin 2x, Á , cos nx, sin nx

L2p

0ƒnsxd sin x dx = pb1 .

2p



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The coefficients of the Fourier series of ƒ are computed using Equations (2), (3), and (4).

So

a0 =

32

, a1 = a2 =Á

= 0,

=

cos kp - 1kp

=

s -1dk- 1

kp.

=1p a c- cos kx

kd

0

p

+ c- 2 cos kxk

dp

2pb

=1p aL

p

0 sin kx dx + L

2p

p

2 sin kx dxb

bk =1pL

2p

0ƒsxd sin kx dx

=1p a csin kx

kd

0

p

+ c2 sin kxkdp

2pb = 0, k Ú 1

=1p aL

p

0 cos kx dx + L

2p

p

2 cos kx dxb

ak =1pL

2p

0ƒsxd cos kx dx

=1

2p aL

p

01 dx + L

2p

p

2 dxb =

32

a0 =1

2pL2p

0ƒsxd dx


x

y

0 � 2�

1

2

(a)

x

y

0 �–�–2� 2� 3� 4�

1

2

(b)

FIGURE 11.16 (a) The step function

(b) The graph of the Fourier series for ƒ is periodic and has the value at each point ofdiscontinuity (Example 1).

3>2ƒsxd = e1, 0 … x … p

2, p 6 x … 2p




and

The Fourier series is

Notice that at where the function ƒ(x) jumps from 1 to 2, all the sine terms vanish,leaving 3 2 as the value of the series. This is not the value of ƒ at since TheFourier series also sums to 3 2 at and In fact, all terms in the Fourier se-ries are periodic, of period and the value of the series at is the same as itsvalue at x. The series we obtained represents the periodic function graphed in Figure11.16b, with domain the entire real line and a pattern that repeats over every interval ofwidth The function jumps discontinuously at and atthese points has value 3 2, the average value of the one-sided limits from each side. Theconvergence of the Fourier series of ƒ is indicated in Figure 11.17.

> x = np, n = 0, ;1, ;2, Á2p .

x + 2p2p ,x = 2p .x = 0> ƒspd = 1.p ,> x = p ,

32

-2p asin x +

sin 3x3

+

sin 5x5 +

Áb .

b1 = -2p, b2 = 0, b3 = -

23p

, b4 = 0, b5 = -2

5p, b6 = 0, Á

1

0

1.5

2

x

y

2��

f

f1

f

f3

f

f5

(a)

0 2��

(b)

1

1.5

2

x

y

0 2��

(c)

x

y

1

1.5

2

f

f9

0x

y

2��

(d)

1

1.5

2 f

f15

0x

y

2��

(e)

1

1.5

2

FIGURE 11.17 The Fourier approximation functions and of the function in Example 1.ƒsxd = e1, 0 … x … p

2, p 6 x … 2pƒ15ƒ1, ƒ3, ƒ5, ƒ9 ,



bounce11.html?5_6

Convergence of Fourier Series

Taylor series are computed from the value of a function and its derivatives at a single pointand cannot reflect the behavior of a discontinuous function such as ƒ in Example 1

past a discontinuity. The reason that a Fourier series can be used to represent such functionsis that the Fourier series of a function depends on the existence of certain integrals, whereasthe Taylor series depends on derivatives of a function near a single point. A function can befairly “rough,” even discontinuous, and still be integrable.

The coefficients used to construct Fourier series are precisely those one should chooseto minimize the integral of the square of the error in approximating ƒ by That is,

is minimized by choosing and as we did. While Taylor seriesare useful to approximate a function and its derivatives near a point, Fourier series mini-mize an error which is distributed over an interval.

We state without proof a result concerning the convergence of Fourier series. A func-tion is piecewise continuous over an interval I if it has finitely many discontinuities on theinterval, and at these discontinuities one-sided limits exist from each side. (See Chapter 5,Additional Exercises 11–18.)

b1, b2, Á , bna0, a1, a2, Á an

L2p

0[ƒsxd - ƒnsxd]2 dx

ƒn .

x = a ,


THEOREM 24 Let ƒ(x) be a function such that ƒ and are piecewise contin-uous on the interval Then ƒ is equal to its Fourier series at all pointswhere ƒ is continuous. At a point c where ƒ has a discontinuity, the Fourier seriesconverges to

where and are the right- and left-hand limits of ƒ at c.ƒsc-dƒsc + d

ƒsc + d + ƒsc-d2

[0, 2p] .ƒ¿




EXERCISES 11.11

Finding Fourier SeriesIn Exercises 1–8, find the Fourier series associated with the givenfunctions. Sketch each function.

1.

2.

3.

4.

5. ƒsxd = ex 0 … x … 2p .

ƒsxd = e x2, 0 … x … p

0, p 6 x … 2p

ƒsxd = e x, 0 … x … p

x - 2p, p 6 x … 2p

ƒsxd = e1, 0 … x … p

-1, p 6 x … 2p

ƒsxd = 1 0 … x … 2p .

6.

7.

8.

Theory and ExamplesEstablish the results in Exercises 9–13, where p and q are positiveintegers.

9. L2p

0 cos px dx = 0 for all p .

ƒsxd = e2, 0 … x … p

-x, p 6 x … 2p

ƒsxd = e cos x, 0 … x … p

0, p 6 x … 2p

ƒsxd = e ex, 0 … x … p

0, p 6 x … 2p



tcu1111a.html

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10.

11.

12.

13.

sHint: sin A cos B = s1>2d[sin sA + Bd + sin sA - Bd].dL

2p

0 sin px cos qx dx = 0 for all p and q .

sHint: sin A sin B = s1>2d[cos sA - Bd - cos sA + Bd].dL

2p

0 sin px sin qx dx = e0, if p Z q

p, if p = q.

sHint: cos A cos B = s1>2d[cossA + Bd + cossA - Bd].dL

2p

0 cos px cos qx dx = e0, if p Z q

p, if p = q.

L2p

0 sin px dx = 0 for all p .

14. Fourier series of sums of functions If ƒ and g both satisfy theconditions of Theorem 24, is the Fourier series of on

the sum of the Fourier series of ƒ and the Fourier series ofg? Give reasons for your answer.

15. Term-by-term differentiation

a. Use Theorem 24 to verify that the Fourier series for inExercise 3 converges to

b. Although show that the series obtained by term-by-term differentiation of the Fourier series in part (a)diverges.

16. Use Theorem 24 to find the Value of the Fourier series determined

in Exercise 4 and show thatp

6

2

= aq

n = 1 1n2 .

ƒ¿sxd = 1,

ƒsxd for 0 6 x 6 2p .ƒsxd

[0, 2p]ƒ + g



tcu1111a.html

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Chapter 11 Additional and Advanced Exercises 843

Convergence or DivergenceWhich of the series defined by the formulas in Exercises 1–4converge, and which diverge? Give reasons for your answers.

1. 2.

3. 4.

Which of the series defined by the formulas in Exercises 5–8converge, and which diverge? Give reasons for your answers.

5.

(Hint: Write out several terms, see which factors cancel, and thengeneralize.)

6.

7.

8. if n is odd, if n is even

Choosing Centers for Taylor SeriesTaylor’s formula

expresses the value of ƒ at x in terms of the values of ƒ and its deriva-tives at In numerical computations, we therefore need a to be apoint where we know the values of ƒ and its derivatives. We also needa to be close enough to the values of ƒ we are interested in to make

so small we can neglect the remainder.In Exercises 9–14, what Taylor series would you choose to repre-

sent the function near the given value of x? (There may be more thanone good answer.) Write out the first four nonzero terms of the seriesyou choose.

9. 10.

11. 12.

13. 14.

Theory and Examples15. Let a and b be constants with Does the sequence

converge? If it does converge, what is the limit?5san+ bnd1>n6

0 6 a 6 b .

tan-1 x near x = 2cos x near x = 69

ln x near x = 1.3ex near x = 0.4

sin x near x = 6.3cos x near x = 1

sx - adn + 1

x = a .

+

ƒsndsadn!

sx - adn+


sx - adn + 1


ƒ–sad2!

sx - ad2+

Á

an = n>3nan = 1>3n

a1 = a2 = 1, an + 1 =

11 + an

if n Ú 2

a1 = a2 = 7, an + 1 =

nsn - 1dsn + 1d

an if n Ú 2

a1 = 1, an + 1 =

nsn + 1dsn + 2dsn + 3d

an

gq

n=1 an

aq

n = 2 logn sn!d

n3aq

n = 1s -1dn tanh n

aq

n = 1 stan-1 nd2

n2+ 1a

q

n = 1

1

s3n - 2dn + s1>2d

gq

n=1 an

Chapter 11 Additional and Advanced Exercises

16. Find the sum of the infinite series

17. Evaluate

18. Find all values of x for which

converges absolutely.

19. Generalizing Euler’s constant The accompanying figureshows the graph of a positive twice-differentiable decreasingfunction ƒ whose second derivative is positive on Foreach n, the number is the area of the lunar region between the curve and the line segment joining the points (n, ƒ(n)) and

a. Use the figure to show that

b. Then show the existence of

c. Then show the existence of

If the limit in part (c) is Euler’s constant(Section 11.3, Exercise 41). (Source: “Convergence with Pic-tures” by P. J. Rippon, American Mathematical Monthly, Vol. 93,No. 6, 1986, pp. 476–478.)

ƒsxd = 1>x ,

limn: q

c an

k= 1 ƒskd - L

n

1ƒsxd dx d .

limn: q

c an

k= 1 ƒskd -

12

sƒs1d + ƒsndd - Ln

1ƒsxd dx d .

gq

n=1 An 6 s1>2dsƒs1d - ƒs2dd .

ƒsn + 1dd .sn + 1,

An

s0, q d .

aq

n = 1

nxn

sn + 1ds2x + 1dn

aq

n = 0 L

n + 1

n

11 + x2 dx .

+

3108 +

7109 +

Á .

1 +

210

+

3102 +

7103 +

2104 +

3105 +

7106 +

2107



20. This exercise refers to the “right side up” equilateral triangle withsides of length 2b in the accompanying figure. “Upside down”equilateral triangles are removed from the original triangle as thesequence of pictures suggests. The sum of the areas removed fromthe original triangle forms an infinite series.

a. Find this infinite series.

b. Find the sum of this infinite series and hence find the totalarea removed from the original triangle.

c. Is every point on the original triangle removed? Explain whyor why not.

21. a. Does the value of

appear to depend on the value of a? If so, how?

b. Does the value of

appear to depend on the value of b? If so, how?

c. Use calculus to confirm your findings in parts (a) and (b).

22. Show that if converges, then

converges.

aq

n = 1 a1 + sin sand

2bn

gq

n=1 an

limn: q

a1 -

cos sa>ndbn

bn

, a and b constant, b Z 0,

limn: q

a1 -

cos sa>ndn bn

, a constant ,

2b

2b 2b

2b

2b 2b

2b

2b 2b • • •

f(4)

1

y � f(x)

0 2 3 4 5

f(3)

f(2)

f(1)

…

A1

A2

A2

A3

A3

…

x

y


T

23. Find a value for the constant b that will make the radius of con-vergence of the power series

equal to 5.

24. How do you know that the functions sin x, ln x, and are notpolynomials? Give reasons for your answer.

25. Find the value of a for which the limit

is finite and evaluate the limit.

26. Find values of a and b for which

27. Raabe’s (or Gauss’s) test The following test, which we statewithout proof, is an extension of the Ratio Test.

Raabe’s test: If is a series of positive constants andthere exist constants C, K, and N such that

(1)

where for then converges if and diverges if

Show that the results of Raabe’s test agree with what youknow about the series and

28. (Continuation of Exercise 27.) Suppose that the terms of are defined recursively by the formulas

Apply Raabe’s test to determine whether the series converges.

29. If converges, and if and for all n,

a. Show that converges.

b. Does converge? Explain.

30. (Continuation of Exercise 29.) If converges, and iffor all n, show that converges.

(Hint: First show that )

31. Nicole Oresme’s Theorem Prove Nicole Oresme’s Theorem that

(Hint: Differentiate both sides of the equation )1 + gq

n=1 xn .1>s1 - xd =

1 +

12

# 2 +

14

# 3 +Á

+

n

2n - 1 +Á

= 4.

ƒ ln s1 - and ƒ … an>s1 - and .

gq

n=1 ln s1 - and1 7 an 7 0gq

n=1 an

gq

n=1 an>s1 - andgq

n=1 an2

an 7 0an Z 1gq

n=1 an

u1 = 1, un + 1 =

s2n - 1d2

s2nds2n + 1d un .

gq

n=1 un

gq

n=1 s1>nd .gq

n=1 s1>n2d

C … 1.C 7 1gq

n=1 unn Ú N ,ƒ ƒsnd ƒ 6 K

unun + 1

= 1 +

Cn +

ƒsnd

n2 ,

gq

n=1 un

limx:0

cos saxd - b

2x2 = -1.

limx:0

sin saxd - sin x - x

x3

ex

aq

n = 2 bnxn

ln n




32. a. Show that

for by differentiating the identity

twice, multiplying the result by x, and then replacing x by1 x.

b. Use part (a) to find the real solution greater than 1 of theequation

33. A fast estimate of As you saw if you did Exercise 127 inSection 11.1, the sequence generated by starting with andapplying the recursion formula convergesrapidly to To explain the speed of the convergence, let

(See the accompanying figure.) Then

Use this equality to show that

34. If is a convergent series of positive numbers, can any-thing be said about the convergence of Givereasons for your answer.

35. Quality control

a. Differentiate the series

to obtain a series for 1>s1 - xd2 .

11 - x

= 1 + x + x2+

Á+ xn

+Á

gq

n=1 ln s1 + and?gq

n=1 an

1

cosxn

0

1

xn

xn

�n

x

y

0 6 Pn + 1 6

16

sPnd3 .

=

13!

APn B3 -

15!

APn B5 +Á .

= Pn - sin Pn

= Pn - cos ap2

- Pnb Pn + 1 =

p

2- xn - cos xn

Pn = sp>2d - xn .p>2.

xn + 1 = xn + cos xn

x0 = 1P/2

x = aq

n = 1 nsn + 1d

xn .

>

aq

n = 1 x

n + 1=

x2

1 - x

ƒ x ƒ 7 1

aq

n = 1 nsn + 1d

xn =

2x2

sx - 1d3

b. In one throw of two dice, the probability of getting a roll of 7is If you throw the dice repeatedly, the probabilitythat a 7 will appear for the first time at the nth throw is

where The expected number ofthrows until a 7 first appears is Find the sum ofthis series.

c. As an engineer applying statistical control to an industrialoperation, you inspect items taken at random from theassembly line. You classify each sampled item as either“good” or “bad.” If the probability of an item’s being good isp and of an item’s being bad is the probabilitythat the first bad item found is the nth one inspected is The average number inspected up to and including the firstbad item found is Evaluate this sum, assuming

36. Expected value Suppose that a random variable X may assumethe values 1, 2, 3, with probabilities where is the probability that X equals Suppose alsothat and that The expected value of X, de-noted by E(X), is the number provided the series con-verges. In each of the following cases, show that and find E(X) if it exists. (Hint: See Exercise 35.)

a. b.

c.

37. Safe and effective dosage The concentration in the blood re-sulting from a single dose of a drug normally decreases with timeas the drug is eliminated from the body. Doses may thereforeneed to be repeated periodically to keep the concentration fromdropping below some particular level. One model for the effect ofrepeated doses gives the residual concentration just before the

dose as

where change in concentration achievable by a singledose (mg mL), elimination constant and between doses (h). See the accompanying figure.

a. Write in closed form as a single fraction, and findR = limn:q Rn .

Rn

t0

C0

0

Time (h)

Con

cent

ratio

n (m

g/m

L)

C1 � C0 � C0e–k t0

R1 � C0e–k t0

R2R3

Rn

Cn�1C2

t

C

t0 = timesh-1d ,k = the>C0 = the

Rn = C0 e-kt0+ C0 e-2k t0

+Á

+ C0 e-nk t0 ,

sn + 1dst

pk =

1ksk + 1d

=

1k

-

1k + 1

pk =

5k - 1

6kpk = 2-k

gq

k=1 pk = 1gq

k=1 kpk ,gq

k=1 pk = 1.pk Ú 0k sk = 1, 2, 3, Á d .

pkp1, p2, p3, Á ,Á ,

0 6 p 6 1.gq

n=1 npn - 1q .

pn - 1q .q = 1 - p ,

gq

n=1 nqn - 1p .q = 1 - p = 5>6.qn - 1p ,

p = 1>6.

T



b. Calculate and for andHow good an estimate of R is

c. If and find the smallest n such that

(Source: Prescribing Safe and Effective Dosage, B. Horelick andS. Koont, COMAP, Inc., Lexington, MA.)

38. Time between drug doses (Continuation of Exercise 37.) If adrug is known to be ineffective below a concentration andharmful above some higher concentration one needs to findvalues of and that will produce a concentration that is safe(not above ) but effective (not below ). See the accompany-ing figure. We therefore want to find values for and forwhich

Thus When these values are substituted in theequation for R obtained in part (a) of Exercise 37, the resultingequation simplifies to

To reach an effective level rapidly, one might administer a “load-ing” dose that would produce a concentration of Thiscould be followed every hours by a dose that raises the concen-tration by

a. Verify the preceding equation for

b. If and the highest safe concentration is e timesthe lowest effective concentration, find the length of timebetween doses that will assure safe and effectiveconcentrations.

c. Given and determine a scheme for administering the drug.

d. Suppose that and that the smallest effectiveconcentration is 0.03 mg mL. A single dose that produces aconcentration of 0.1 mg mL is administered. About how longwill the drug remain effective?

39. An infinite product The infinite product

qq

n = 1s1 + and = s1 + a1ds1 + a2ds1 + a3d Á

>>k = 0.2 h-1

k = 0.02 h-1 ,CH = 2 mg>mL, CL = 0.5 mg/mL,

k = 0.05 h-1

t0 .

C0 = CH - CL mg>mL.t0

CH mg>mL.

t0 =

1k

ln CH

CL.

C0 = CH - CL .

t0

CL

0 Time

Con

cent

ratio

n in

blo

od

C0

Highest safe levelCH

Lowest effective level

t

C

R = CL and C0 + R = CH .

t0C0

CLCH

t0C0

CH ,CL

Rn 7 s1>2dR .t0 = 10 h ,k = 0.01 h-1

R10 ?t0 = 10 h .C0 = 1 mg>mL, k = 0.1 h-1 ,R10R1 is said to converge if the series

obtained by taking the natural logarithm of the product, con-verges. Prove that the product converges if for every nand if converges. (Hint: Show that

when )

40. If p is a constant, show that the series

a. converges if b. diverges if In general, ifand n takes on the values

we find that andso on. If then

converges if and diverges if

41. a. Prove the following theorem: If is a sequence of numberssuch that every sum is bounded, then the series

converges and is equal to

Outline of proof: Replace by and by

for If show that

Because for some constant M, the series

converges absolutely and has a limit as

Finally, if then approaches zero as because

Hence the sequence ofpartial sums of the series converges and the limit isgq

k=1 tk>sk sk + 1dd .gck>k

ƒ c2n + 1 ƒ = ƒ t2n + 1 - t2n ƒ 6 2M .n : qc2n + 1>s2n + 1d

s2n + 1 - s2n =s2n = g2nk=1 ck>k ,

n : q .s2n + 1

aq

k = 1

tkk sk + 1d

ƒ tk ƒ 6 M

= a2n

k = 1

tkk sk + 1d

+

t2n + 1

2n + 1.

+Á

+ t2n a 12n

-

12n + 1

b +

t2n + 1

2n + 1

s2n + 1 = t1 a1 -

12b + t2 a1

2-

13b

s2n + 1 = g2n+1k=1 ck>k ,n Ú 2.

tn - tn - 1cnt1c1

gq

n=1 tn>(n(n + 1)) .gq

n=1 cn>ntn = gn

k=1 ck

5cn6p … 1.p 7 1

Lq

a

dxƒ1sxdƒ2sxd Á ƒnsxdsƒn + 1sxddp

ƒnsad 7 1,ƒ2sxd = ln x, ƒ3sxd = ln sln xd ,1, 2, 3, Á ,

ƒ1sxd = x, ƒn + 1sxd = ln sƒnsxdd ,p … 1.p 7 1,

1 + aq

n = 3

1n # ln n # [ln sln nd]p

ƒ an ƒ 6 1>2.

ƒ ln s1 + and ƒ …

ƒ an ƒ

1 - ƒ an ƒ

… 2 ƒ an ƒ

gq

n=1 ƒ an ƒ

an 7 -1

aq

n = 1 ln s1 + and ,





b. Show how the foregoing theorem applies to the alternatingharmonic series

c. Show that the series

converges. (After the first term, the signs are two negative,two positive, two negative, two positive, and so on in thatpattern.)

42. The convergence of for

a. Show by long division or otherwise that

b. By integrating the equation of part (a) with respect to t from 0to x, show that

where

Rn + 1 = s -1dn + 1

Lx

0

t n + 1

1 + t dt .

+ s -1dn xn + 1

n + 1+ Rn + 1

ln s1 + xd = x -

x2

2+

x3

3-

x4

4+

Á

11 + t

= 1 - t + t2- t3

+Á

+ s -1dnt n+

s -1dn + 1t n + 1

1 + t.

-1 6 x … 1gq

n=1 [s -1dn - 1xn]>n to ln s1 + xd

1 -

12

-

13

+

14

+

15

-

16

-

17

+Á .

1 -

12

+

13

-

14

+

15

-

16

+Á .

c. If show that

Hint: As t varies from 0 to x,

and

d. If show that

Hint: If then and

e. Use the foregoing results to prove that the series

converges to for -1 6 x … 1.ln s1 + xd

x -

x2

2+

x3

3-

x4

4+

Á+

s -1dnxn + 1

n + 1+

Á

` t n + 1

1 + t` …

ƒ t ƒn + 1

1 - ƒ x ƒ

.b

ƒ 1 + t ƒ Ú 1 - ƒ x ƒx 6 t … 0,a

ƒRn + 1 ƒ … `Lx

0

t n + 1

1 - ƒ x ƒ

dt ` =

ƒ x ƒn + 2

sn + 2ds1 - ƒ x ƒd.

-1 6 x 6 0,

`Lx

0ƒstd dt ` … L

x

0ƒƒstd ƒ dt.b

1 + t Ú 1 and t n + 1>s1 + td … t n + 1 ,

a

ƒ Rn + 1 ƒ … Lx

0t n + 1 dt =

xn + 2

n + 2.

x Ú 0,




Chapter 11 Practice Exercises

Convergent or Divergent SequencesWhich of the sequences whose nth terms appear in Exercises 1–18converge, and which diverge? Find the limit of each convergent se-quence.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

Convergent SeriesFind the sums of the series in Exercises 19–24.

19. 20.

21. 22.

23. 24. aq

n = 1s -1dn

34na

q

n = 0e-n

aq

n = 3

-8s4n - 3ds4n + 1da

q

n = 1

9s3n - 1ds3n + 2d

aq

n = 2

-2nsn + 1da

q

n = 3

1s2n - 3ds2n - 1d

an =

s -4dn

n!an =

sn + 1d!n!

an = 2n 2n + 1an = ns21>n- 1d

an = a3n b1>n

an = An 3n

n

an = a1 +

1n b

-n

an = an - 5n bn

an =

ln s2n3+ 1d

nan =

n + ln nn

an =

ln s2n + 1dnan =

ln sn2dn

an = sin npan = sin np2

an = 1 + s0.9dnan =

1 - 2n

2n

an =

1 - s -1dn2nan = 1 +

s -1dn

n

Convergent or Divergent SeriesWhich of the series in Exercises 25–40 converge absolutely, whichconverge conditionally, and which diverge? Give reasons for youranswers.

25. 26. 27.

28. 29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

Power SeriesIn Exercises 41–50, (a) find the series’ radius and interval of conver-gence. Then identify the values of x for which the series converges (b)absolutely and (c) conditionally.

41. 42.

43. 44.

45. 46. aq

n = 1

xn2naq

n = 1 xn

nn

aq

n = 0 sn + 1ds2x + 1dn

s2n + 1d2naq

n = 1 s -1dn - 1s3x - 1dn

n2

aq

n = 1 sx - 1d2n - 2

s2n - 1d!aq

n = 1 sx + 4dn

n3n

aq

n = 2

1

n2n2- 1

aq

n = 1

12nsn + 1dsn + 2d

aq

n = 1 2n 3n

nnaq

n = 1 s -3dn

n!

aq

n = 1 s -1dnsn2

+ 1d2n2

+ n - 1aq

n = 1 n + 1

n!

aq

n = 1 s -1dn 3n2

n3+ 1a

q

n = 1

s -1dn

n2n2+ 1

aq

n = 3

ln nln sln nda

q

n = 1 ln n

n3

aq

n = 2

1n sln nd2a

q

n = 1

s -1dn

ln sn + 1daq

n = 1

12n3

aq

n = 1 s -1dn2n

aq

n = 1 -5na

q

n = 1

12n



Chapter 11 Practice Exercises 841

47. 48.

49. 50.

Maclaurin SeriesEach of the series in Exercises 51–56 is the value of the Taylor seriesat of a function ƒ(x) at a particular point. What function andwhat point? What is the sum of the series?

51.

52.

53.

54.

55.

56.

Find Taylor series at for the functions in Exercises 57–64.

57. 58.

59. 60.

61. 62.

63. 64.

Taylor SeriesIn Exercises 65–68, find the first four nonzero terms of the Taylorseries generated by ƒ at

65.

66.

67.

68.

Initial Value ProblemsUse power series to solve the initial value problems in Exercises 69–76.

69. 70.

71. 72.

73. 74.

75. 76. y¿ - y = -x, ys0d = 2y¿ - y = x, ys0d = 1

y¿ + y = x, ys0d = 0y¿ - y = 3x, ys0d = -1

y¿ + y = 1, ys0d = 0y¿ + 2y = 0, ys0d = 3

y¿ - y = 0, ys0d = -3y¿ + y = 0, ys0d = -1

ƒsxd = 1>x at x = a 7 0

ƒsxd = 1>sx + 1d at x = 3

ƒsxd = 1>s1 - xd at x = 2

ƒsxd = 23 + x2 at x = -1

x = a .

e-x2

e spx>2d

cos 25xcos sx5>2d

sin 2x3

sin px

11 + x3

11 - 2x

x = 0

+ s -1dn - 1 1

s2n - 1d A23 B2n - 1+

Á

123-

1

923+

1

4523-

Á

1 + ln 2 +

sln 2d2

2!+

Á+

sln 2dn

n!+

Á

1 -

p2

9 # 2!+

p4

81 # 4!-

Á+ s -1dn

p2n

32ns2nd!+

Á

p -

p3

3!+

p5

5!-

Á+ s -1dn

p2n + 1

s2n + 1d!+

Á

23

-

418

+

881

-Á

+ s -1dn - 1 2n

n3n +Á

1 -

14

+

116

-Á

+ s -1dn 14n +

Á

x = 0

aq

n = 1scoth ndxna

q

n = 1scsch ndxn

aq

n = 0 s -1dnsx - 1d2n + 1

2n + 1aq

n = 0 sn + 1dx2n - 1

3nNonelementary IntegralsUse series to approximate the values of the integrals in Exercises77–80 with an error of magnitude less than (The answer sectiongives the integrals’ values rounded to 10 decimal places.)

77. 78.

79. 80.

Indeterminate FormsIn Exercises 81–86:

a. Use power series to evaluate the limit.

b. Then use a grapher to support your calculation.

81. 82.

83. 84.

85. 86.

87. Use a series representation of sin 3x to find values of r and s forwhich

88. a. Show that the approximation in Section11.10, Example 9, leads to the approximation

b. Compare the accuracies of the approximations and by comparing the graphs of

and Describe what you find.

Theory and Examples89. a. Show that the series

converges.

b. Estimate the magnitude of the error involved in using the sumof the sines through to approximate the sum of theseries. Is the approximation too large, or too small? Givereasons for your answer.

90. a. Show that the series converges.

b. Estimate the magnitude of the error in using the sum of thetangents through to approximate the sum of theseries. Is the approximation too large, or too small? Givereasons for your answer.

- tan s1>41d

aq

n = 1 atan

12n

- tan 1

2n + 1b

n = 20

aq

n = 1 asin

12n

- sin 1

2n + 1b

gsxd = sin x - s6x>s6 + x2dd .ƒsxd = sin x - xsin x L 6x>s6 + x2d

sin x L x

6x>s6 + x2d .sin x L

csc x L 1>x + x>6limx:0

asin 3x

x3 +

rx2 + sb = 0.

limy:0

y2

cos y - cosh ylimz :0

1 - cos2 z

ln s1 - zd + sin z

limh:0

ssin hd>h - cos h

h2limt:0

a 12 - 2 cos t

-

1t2 b

limu:0

eu - e-u

- 2uu - sin u

limx:0

7 sin x

e2x- 1

L1>64

0 tan-1 x2x

dxL1>2

0 tan-1 x

x dx

L1

0x sin sx3d dxL

1>20

e-x3

dx

10-8 .

T

T

T

T



91. Find the radius of convergence of the series

92. Find the radius of convergence of the series

93. Find a closed-form formula for the nth partial sum of the seriesand use it to determine the convergence

or divergence of the series.

94. Evaluate by finding the limits as ofthe series’ nth partial sum.

95. a. Find the interval of convergence of the series

b. Show that the function defined by the series satisfies a differ-ential equation of the form

and find the values of the constants a and b.

96. a. Find the Maclaurin series for the function

b. Does the series converge at Explain.

97. If and are convergent series of nonnegativenumbers, can anything be said about Give reasonsfor your answer.

98. If and are divergent series of nonnegativenumbers, can anything be said about Give reasonsfor your answer.

99. Prove that the sequence and the series both converge or both diverge.

100. Prove that converges if for all n andconverges.

101. (Continuation of Section 4.7, Exercise 27.) If you did Exercise27 in Section 4.7, you saw that in practice Newton’s methodstopped too far from the root of to give a use-ful estimate of its value, Prove that nevertheless, for anystarting value the sequence of ap-proximations generated by Newton’s method really does con-verge to 1.

102. Suppose that are positive numbers satisfyingthe following conditions:

i.

ii. the series diverges.a2 + a4 + a8 + a16 +Á

a1 Ú a2 Ú a3 ÚÁ ;

a1, a2, a3, Á , an

x0, x1, x2, Á , xn, Áx0 Z 1,x = 1.

ƒsxd = sx - 1d40

gq

n=1 an

an 7 0gq

n=1 san>s1 + andd

gq

k=1 sxk+ 1 - xkd5xn6gq

n=1 an bn ?gq

n=1 bngq

n=1 an

gq

n=1 an bn ?gq

n=1 bngq

n=1 an

x = 1?

x2>s1 + xd .

d2y

dx2 = xay + b

+

1 # 4 # 7 # Á # s3n - 2ds3nd!

x3n+

Á .

y = 1 +

16

x3+

1180

x6+

Á

n : qgq

k=2 s1>sk2- 1dd

gq

n=2 ln s1 - s1>n2dd

aq

n = 1

3 # 5 # 7 # Á # s2n + 1d4 # 9 # 14 # Á # s5n - 1d

sx - 1dn .

aq

n = 1 2 # 5 # 8 # Á # s3n - 1d

2 # 4 # 6 # Á # s2nd xn .

Show that the series

diverges.

103. Use the result in Exercise 102 to show that

diverges.

104. Suppose you wish to obtain a quick estimate for the value of

There are several ways to do this.

a. Use the Trapezoidal Rule with to estimate

b. Write out the first three nonzero terms of the Taylor series atfor to obtain the fourth Taylor polynomial P(x)

for Use to obtain another estimate for

c. The second derivative of is positive for allExplain why this enables you to conclude that the

Trapezoidal Rule estimate obtained in part (a) is too large.(Hint: What does the second derivative tell you about thegraph of a function? How does this relate to the trapezoidalapproximation of the area under this graph?)

d. All the derivatives of are positive for Ex-plain why this enables you to conclude that all Maclaurinpolynomial approximations to ƒ(x) for x in [0, 1] will be toosmall. (Hint: )

e. Use integration by parts to evaluate

Fourier SeriesFind the Fourier series for the functions in Exercises 105–108. Sketcheach function.

105.

106.

107.

108. ƒsxd = ƒ sin x ƒ , 0 … x … 2p

ƒsxd = ep - x, 0 … x … p

x - 2p, p 6 x … 2p

ƒsxd = e x, 0 … x … p

1, p 6 x … 2p

ƒsxd = e0, 0 … x … p

1, p 6 x … 2p

110

x2ex dx .

ƒsxd = Pnsxd + Rnsxd .

x 7 0.ƒsxd = x2ex

x 7 0.ƒsxd = x2ex

110

x2ex dx .11

0 Psxd dxx2ex .

x2exx = 0

110

x2ex dx .n = 211

0 x2ex dx .

1 + aq

n = 2

1n ln n

a1

1+

a2

2+

a3

3+

Á




839

Chapter 11 Questions to Guide Your Review

1. What is an infinite sequence? What does it mean for such a se-quence to converge? To diverge? Give examples.

2. What is a nondecreasing sequence? Under what circumstancesdoes such a sequence have a limit? Give examples.

3. What theorems are available for calculating limits of sequences?Give examples.

4. What theorem sometimes enables us to use l’Hôpital’s Rule tocalculate the limit of a sequence? Give an example.

5. What six sequence limits are likely to arise when you work withsequences and series?

6. What is an infinite series? What does it mean for such a series toconverge? To diverge? Give examples.

7. What is a geometric series? When does such a series converge?Diverge? When it does converge, what is its sum? Give examples.

8. Besides geometric series, what other convergent and divergent se-ries do you know?

9. What is the nth-Term Test for Divergence? What is the idea be-hind the test?

10. What can be said about term-by-term sums and differences ofconvergent series? About constant multiples of convergent and di-vergent series?

11. What happens if you add a finite number of terms to a convergentseries? A divergent series? What happens if you delete a finitenumber of terms from a convergent series? A divergent series?

12. How do you reindex a series? Why might you want to do this?

13. Under what circumstances will an infinite series of nonnegativeterms converge? Diverge? Why study series of nonnegativeterms?

14. What is the Integral Test? What is the reasoning behind it? Givean example of its use.

15. When do p-series converge? Diverge? How do you know? Giveexamples of convergent and divergent p-series.

16. What are the Direct Comparison Test and the Limit ComparisonTest? What is the reasoning behind these tests? Give examples oftheir use.

17. What are the Ratio and Root Tests? Do they always give you theinformation you need to determine convergence or divergence?Give examples.

18. What is an alternating series? What theorem is available for deter-mining the convergence of such a series?

19. How can you estimate the error involved in approximating thesum of an alternating series with one of the series’ partial sums?What is the reasoning behind the estimate?

20. What is absolute convergence? Conditional convergence? Howare the two related?

21. What do you know about rearranging the terms of an absolutelyconvergent series? Of a conditionally convergent series? Giveexamples.

22. What is a power series? How do you test a power series for con-vergence? What are the possible outcomes?

23. What are the basic facts about

a. term-by-term differentiation of power series?

b. term-by-term integration of power series?

c. multiplication of power series?

Give examples.

24. What is the Taylor series generated by a function ƒ(x) at a pointWhat information do you need about ƒ to construct the

series? Give an example.

25. What is a Maclaurin series?

x = a?


Questions to Guide Your ReviewChapter 11



26. Does a Taylor series always converge to its generating function?Explain.

27. What are Taylor polynomials? Of what use are they?

28. What is Taylor’s formula? What does it say about the errors in-volved in using Taylor polynomials to approximate functions? Inparticular, what does Taylor’s formula say about the error in a lin-earization? A quadratic approximation?

29. What is the binomial series? On what interval does it converge?How is it used?

30. How can you sometimes use power series to solve initial valueproblems?

31. How can you sometimes use power series to estimate the valuesof nonelementary definite integrals?

32. What are the Taylor series for and How do

you estimate the errors involved in replacing these series withtheir partial sums?

33. What is a Fourier series? How do you calculate the Fourier coeffi-cients and for a function ƒ(x) defined onthe interval

34. State the theorem on convergence of the Fourier series for ƒ(x)when ƒ and are piecewise continuous on [0, 2p] .ƒ¿

[0, 2p]?b1, b2, Áa0, a1, a2, Á

tan-1 x?cos x, ln s1 + xd, ln [s1 + xd>s1 - xd] ,1>s1 - xd, 1>s1 + xd, ex, sin x,



Chapter 11 847

Chapter 11 Technology Application Projects

Mathematica Maple ModuleBouncing BallThe model predicts the height of a bouncing ball, and the time until it stops bouncing.

Mathematica Maple ModuleTaylor Polynomial Approximations of a FunctionA graphical animation shows the convergence of the Taylor polynomials to functions having derivatives of all orders over an interval in theirdomains.

/

/


Technology Application Projects


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